I) Passive picture. The einbein $e$ is not an invariant but transforms as
$$ e~=~e^{\prime} \frac{d\tau^{\prime}}{d\tau}\tag{1} $$
under a reparametrization of the world-line (WL) parameter
$$ \tau\longrightarrow \tau^{\prime}=f(\tau).\tag{2} $$
In other words, $\omega:= e \mathrm{d}\tau\in \Gamma(T^{\ast}I) $
is a one-form on the 1-dimensional WL manifold $I$. The particle position
$$ x^{\mu}~=~x^{\prime \mu}\tag{3} $$
is invariant, while the particle velocity transforms as
$$ \dot{x}^{\mu}~=~\dot{x}^{\prime \mu}\frac{d\tau^{\prime}}{d\tau}.\tag{4}$$
These transformation rules (1)-(4) can be seen in many ways. One way is that the action
$$ S~=~\int \! \mathrm{d}\tau ~L , \qquad L~:=~\frac{\dot{x}^2}{2e}-\frac{e m^2}{2},\tag{5}$$
should be invariant under reparametrizations (2). See also this related Phys.SE post.
Let us call variations in the WL manifold $I$ for horizontal and variations in the target space (TS) for vertical. Then the infinitesimal horisontal variation
$$\delta_h\tau ~:=~\tau^{\prime} - \tau ~=~-\eta, \tag{6}$$
where $\eta$ is an infiniteimal parameter. More generally, the infinitesimal horisontal variation is
$$ \delta_h~\stackrel{(6)}{=}~ -\eta\frac{d}{d\tau}.\tag{7}$$
The total infinitesimal variations are
$$\begin{align} \delta x~:=~&x^{\prime}(\tau^{\prime})-x(\tau)~\stackrel{(3)}{=}~0, \cr
\delta e~:=~&e^{\prime}(\tau^{\prime})-e(\tau)~\stackrel{(1)+(6)}{=}~e\frac{d\eta}{d\tau} .
\end{align}\tag{8}$$
The infinitesimal vertical variations
$$\delta_v~=~\delta-\delta_h\tag{9}$$
are
$$\begin{align} \delta_v x~:=~&x^{\prime}(\tau)-x(\tau)
~\stackrel{(7)+(8)+(9)}{=}~\eta\frac{d x^{\mu}}{d\tau}, \cr
\delta_v e~:=~&e^{\prime}(\tau)-e(\tau)
~\stackrel{(7)+(8)+(9)}{=}~\frac{d}{d\tau}(\eta e) .
\end{align}\tag{10}$$
II) Active picture. From the perspective of the 1-dimensional WL manifold $I$, the infinitesimal transformation can e.g. be encoded via Lie derivatives ${\cal L}_Y$ wrt. a vector field
$$ Y ~=~\eta \frac{d}{d\tau}~\in~ \Gamma(TI) \tag{11} $$
on the 1-dimensional WL manifold $I$. The Lie derivatives are
$$ {\cal L}_Y x^{\mu}~=~Y[x^{\mu}]~=~\eta \frac{dx^{\mu}}{d\tau},\tag{12} $$
$$\begin{align} ({\cal L}_Ye)\mathrm{d}\tau~:=~&{\cal L}_Y\omega
~=~\{\mathrm{d}, i_Y\}\omega~=~\mathrm{d}i_Y\omega \cr
~=~&\mathrm{d}(i_Y\omega)~=~\mathrm{d}(\eta e)
~=~\mathrm{d}\tau\frac{d}{d\tau}(\eta e),\end{align} \tag{13} $$
and hence
$$ {\cal L}_Ye ~\stackrel{(8)}{=}~\frac{d}{d\tau}(\eta e).\tag{14} $$
The above should be compared with eq. (1.10) in Ref. 1
$$ \tau\to \tau^{\prime}~=~\tau-\eta, \qquad
\delta_v x^{\mu}~=~\eta\frac{d x^{\mu}}{d\tau}, \qquad \delta_v e ~=~\frac{d}{d\tau}(\eta e), \tag{1.10}$$
respectively.
III) Classical BV formulation. Let us mention for completeness that the gauge transformation $\delta$ can be encoded as a BRST transformation, cf. e.g. Ref. 2 and this Phys.SE post. Roughly speaking, the Grassmann-even gauge parameter $\eta$ is then replaced by a Grassmann-odd Faddeev-Popov (FP) ghost $C$. (Actually, the gauge parameter $\eta$ will more precisely be replaced with the combination $e^{1-r}C$, where $r\in\mathbb{R}$ is a power, to be more general, cf. eq. (26) below.) To minimize the appearances of time derivatives, instead of using the Lagrangian (5), it becomes a bit simpler to start from the Hamiltonian Lagrangian
$$L_H~:=~ p_{\mu} \dot{x}^{\mu} - H, \qquad H~:=~ eT, \qquad T~:=~\frac{1}{2}(p^2+m^2), \qquad p^2~:=~ g^{\mu\nu}(x)~p_{\mu} p_{\nu}, \tag{21} $$
cf. e.g. this Phys.SE post.
Here we will use the Batalin-Vilkovisky (BV) formalism, cf. Ref. 3. The fields
$$ \phi^{\alpha} ~=~ \{ x^{\mu};~p_{\mu};~ e;~C;~ \bar{C};~B\} \tag{22}$$
are positions $x^{\mu}$; momenta $p_{\mu}$; einbein $e$; FP ghost $C$; FP antighost $\bar{C}$; and Lautrup-Nakanishi (LN) Lagrange multiplier $B$, respectively. They are WL tensors of contravariant orders $0$; $0$; $-1$; $r$; $0$; and $0$, respectively. Each field $\phi^{\alpha}$ has a corresponding antifield $\phi^{\ast}_{\alpha}$ of opposite Grassmann parity. The corresponding BV action$^1$
$$\begin{align} S_{BV}~=~&\int \! \mathrm{d}\tau ~L_{BV} , \cr
L_{BV}~=~&L_H +\left(x^{\ast}_{\mu} \dot{x}^{\mu}+p_{\ast}^{\mu} \dot{p}_{\mu} +r C^{\ast}\dot{C} \right)e^{r-1}C +\underbrace{e^r C\dot{e}^{\ast}}_{\sim~e^{\ast}\frac{d}{d\tau}( e^r C)} + B \bar{C}^{\ast},\end{align} \tag{23} $$
satisfies the classical master equation
$$ (S_{BV},S_{BV})~=~0, \tag{24}$$
with antibracket $(\cdot,\cdot)$ on Darboux-form, i.e. the non-zero fundamental antibrackets read
$$ (\phi^{\alpha}(\tau),\phi^{\ast}_{\beta}(\tau^{\prime}))
~=~\delta^{\alpha}_{\beta}~\delta(\tau\!-\!\tau^{\prime}). \tag{25}$$
The Grassmann-odd nilpotent BRST transformation ${\bf s}~=~(S_{BV},\cdot)$ reads
$$\begin{align} {\bf s}x^{\mu}~=~&e^{r-1} C \dot{x}^{\mu}, \qquad
{\bf s}p_{\mu}~=~e^{r-1} C \dot{p}_{\mu}, \qquad
{\bf s}e~=~ \frac{d}{d\tau}( e^r C), \cr
{\bf s}C~=~& re^{r-1} C\dot{C},\qquad
{\bf s}\bar{C}~=~ - B,\qquad
{\bf s}B ~=~0, \end{align} \tag{26} $$
which should be compared with eq. (1.10). The BV gauge-fixing fermion $\psi$ can be chosen on the form
$$ \psi ~:=~\int \! \mathrm{d}\tau~\bar{C}\left(\frac{\xi\rho}{2}B +\chi(e) +\epsilon \frac{d(e/\rho)}{d\tau}\right), \tag{27} $$
where $\xi,\epsilon\in\mathbb{R}$ are gauge-fixing parameters, and $\rho$ is a fiducial einbein. Moreover, $\chi(e)=(e\!-\!e_0)\chi^{\prime}$ is a gauge-fixing condition (which we will assume is affine in $e$, so that the derivative $\chi^{\prime}$ is constant). The gauge-fixed Lagrangian becomes
$$\begin{align} L_{\rm gf}~=~ &\left. L_{BV} \right|_{\phi^{\ast}
~=~\frac{\delta \psi}{\delta \phi}}\cr
~=~& L_H + \overbrace{\underbrace{\left(\chi^{\prime}\bar{C}-\frac{\epsilon}{\rho}\dot{\bar{C}}\right) \frac{d}{d\tau}(e^r C)}_{
\sim~ \bar{C} \left(\chi^{\prime}+\frac{d}{d\tau}\frac{\epsilon}{\rho}\right)\frac{d}{d\tau}(e^r C)
~\sim~ e^r C\left(\chi^{\prime}-\frac{d}{d\tau}\frac{\epsilon}{\rho}\right)\frac{d}{d\tau}\bar{C}
}}^{\text{Faddeev-Popov term}} + \overbrace{B \left(\frac{\xi\rho}{2}B +\chi(e) +\epsilon \frac{d(e/\rho)}{d\tau}\right)}^{\text{gauge-fixing term}}, \end{align} \tag{28} $$
where the $\sim$ symbol means equality up to total time derivative terms.
The physical quantities do not depend on the choice of the gauge-fixing fermion $\psi$, as long as certain rank conditions are met.
IV) Quantum master equation. The odd Laplacian
$$ \Delta~=~(-1)^{|\alpha|}\int\! \mathrm{d}\tau~
\frac{\delta_L}{\delta\phi^{\alpha}(\tau)}
\frac{\delta_L}{\delta\phi^{\ast}_{\alpha}(\tau)}
~=~(-1)^{|\alpha|}\iint\! \mathrm{d}\tau~\mathrm{d}\tau^{\prime}~
\delta(\tau\!-\!\tau^{\prime})\frac{\delta_L}{\delta\phi^{\alpha}(\tau)}
\frac{\delta_L}{\delta\phi^{\ast}_{\alpha}(\tau^{\prime})} \tag{29} $$
is a singular object, which strictly speaking needs to be regularized. We calculate formally
$$ \Delta S_{BV}~\stackrel{(23)+(29)}{=}~
2(n\!-\!r)\iint\! \mathrm{d}\tau~\mathrm{d}\tau^{\prime}~
e(\tau)^{r-1}C(\tau)~
\delta(\tau\!-\!\tau^{\prime}) \frac{d}{d\tau}\delta(\tau\!-\!\tau^{\prime})
$$ $$+r \iint\! \mathrm{d}\tau~\mathrm{d}\tau^{\prime}~
e(\tau)^{r-1}\dot{C}(\tau)~\delta(\tau\!-\!\tau^{\prime})^2~\neq~0,
\tag{30} $$
where $n$ is the target space (TS) dimension. This shows that the BV action (23) does not satisfy the quantum master equation; only the classical master equation. We will discuss appropriate modifications of the BV action (23) in Section VII.
V) Classical BFV formulation. We identify the einbein $e\equiv \rho\lambda $, where $\lambda$ is a Lagrange multiplier. We identify $p_{\lambda}\approx\epsilon B$ with the canonical momentum of the Lagrange multiplier $\lambda$, and we identify the antifield $e^{\ast}\equiv \bar{P}$ with the FP ghost momentum. Introduce an ultra-local Poisson bracket $\{\cdot,\cdot\}_{PB}$ with the following canonical pairs
$$\begin{align} \{x^{\mu}(\tau), p_{\nu}(\tau^{\prime})\}_{PB}
~=~&\delta^{\mu}_{\nu}\frac{1}{\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}), \qquad
\{e(\tau)^rC(\tau), \bar{P}(\tau^{\prime})\}_{PB}
~=~\frac{1}{\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}), \cr
\{\lambda(\tau), B (\tau^{\prime})\}_{PB}
~=~&\frac{1}{\epsilon\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}), \qquad
\{\bar{C}(\tau), P(\tau^{\prime})\}_{PB}
~=~\frac{1}{\epsilon\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}).\end{align}\tag{31} $$
Note the non-Darboux form
$$ \{C(\tau), \bar{P}(\tau^{\prime})\}_{PB}
~=~\frac{e(\tau)^{-r}}{\rho(\tau)}\delta(\tau\!-\!\tau^{\prime}), \qquad
\{ B (\tau), C(\tau^{\prime})\}_{PB}
~=~\frac{r}{\epsilon} \frac{C(\tau)}{e(\tau)} \delta(\tau\!-\!\tau^{\prime}), \tag{32} $$
to ensure that
$$ \{e(\tau)^r C(\tau), B(\tau^{\prime})\}_{PB}~=~0. \tag{33} $$
The BRST transformation ${\bf s}~=~\{\mathbb{Q},\cdot\}_{PB}$ (which is independent of the $\epsilon$-parameter) reads
$$\begin{align}{\bf s}x^{\mu}~=~&e^r C g^{\mu\nu}(x)p_{\nu}
~\approx~ e^{r-1} C \dot{x}^{\mu}, \cr
{\bf s}p_{\mu}
~=~& -\frac{1}{2}e^r C \partial_{\mu}g^{\nu\lambda}(x)~p_{\nu}p_{\lambda}
~\approx~ e^{r-1} C \dot{p}_{\mu}, \cr
{\bf s}e~=~&\rho P~\approx~ \frac{d}{d\tau}( e^r C) , \qquad
{\bf s}C~=~r\frac{C}{e}\rho P ~\approx~ re^{r-1} C\dot{C},\cr
{\bf s}\bar{C}~=~& - B,\qquad
{\bf s} B ~=~0, \end{align} \tag{34} $$
which should be compared with eq. (26). Here the $\approx$ symbol means equality modulo eqs. of motion. The BRST transformation (34) is generated by
$$ \mathbb{Q}~:=~ \int \! \rho \mathrm{d}\tau ~Q, \qquad \{\mathbb{Q}, \mathbb{Q}\}_{PB}~=~0,\tag{35}$$
where
$$ -Q~:=~T e^r C + \epsilon B P ~\approx~ T e^r C + \epsilon B \frac{d}{d\tau}( e^r C)\tag{36}$$
is the BRST charge. The BFV action becomes
$$ S_{BFV}
~=~ \int \! \mathrm{d}\tau~\left(\dot{x}^{\mu}p_{\mu}+e^r C\dot{\bar{P}} \right)
-\left\{ \psi, \mathbb{Q} \right\}_{PB}
~=~ \int \! \mathrm{d}\tau ~L_{BFV} , \tag{37} $$
where the BFV gauge-fixing fermion $\psi$ is
$$ \psi ~:=~\int \! \mathrm{d}\tau \left(\bar{C} \left(\frac{\xi\rho}{2}B +\chi(e) +\epsilon \dot{\lambda}\right) -\bar{P}e\right),\tag{38} $$
and where the BFV Lagrangian reads$^2$
$$\begin{align} L_{BFV}~=~&\left(p_{\mu}\dot{x}^{\mu}+ e^r C\dot{\bar{P}} \right)
+ \epsilon\left( B \dot{\lambda} + \bar{C} \dot{P}\right)
+ \left(-eT +\bar{C}\rho\chi^{\prime} P +B \left(\frac{\xi\rho}{2}B+\chi(e)\right) -\rho\bar{P}P \right)\cr
~\sim~& L_H+ \underbrace{\epsilon\left( B \dot{\lambda} + \bar{C} \dot{P}\right)}_{\text{kinetic term}}+ \bar{P}\left( \frac{d}{d\tau}( e^r C)-\rho P\right) + \underbrace{\bar{C} \rho\chi^{\prime} P}_{\text{FP term}} + \underbrace{B \left(\frac{\xi\rho}{2}B+\chi(e)\right)}_{\text{gauge-fixing term}} \end{align} \tag{39} .$$
VI) Dirac bracket. Let us integrate out the two FP momenta $P$ and $\bar{P}$.
Then the BFV Lagrangian (39) becomes the gauge-fixed Lagrangian (28) from Section III. The corresponding two 2nd class constraints
$$ \Theta~:=~ \rho P - \frac{d}{d\tau}( e^r C)~\approx~0, \qquad \bar{\Theta}~:=~ \rho\bar{P} - \rho\chi^{\prime} \bar{C}+\epsilon\dot{\bar{C}}~\approx~0,\tag{40} $$
has non-zero Poisson bracket
$$ \Delta(\tau,\tau^{\prime} )
~:=~ \{\Theta(\tau), \bar{\Theta}(\tau^{\prime}) \}_{PB}
~=~ -\left(\frac{\rho(\tau) \chi^{\prime}}{\epsilon}+2\frac{d}{d\tau} \right) \delta(\tau\!-\!\tau^{\prime}),\tag{41} $$
with inverse
$$ \Delta^{-1}(\tau,\tau^{\prime} )
~=~ - \frac{1}{4} \exp\left[\frac{R(\tau^{\prime},\tau)\chi^{\prime}}{2\epsilon}\right] {\rm sgn}(\tau\!-\!\tau^{\prime}),
\qquad R(\tau^{\prime},\tau)~:= \int_{\tau}^{\tau^{\prime}}\!\mathrm{d}\tau^{\prime\prime}\rho(\tau^{\prime\prime}). \tag{42} $$
Therefore the Dirac bracket becomes
$$ \{e(\tau)^rC(\tau), \bar{C}(\tau^{\prime})\}_{DB}
~=~ \frac{1}{4\epsilon} \exp\left[\frac{R(\tau^{\prime},\tau)\chi^{\prime}}{2\epsilon}\right] {\rm sgn}(\tau\!-\!\tau^{\prime}).\tag{43} $$
Due to double time derivatives in the FP term in the gauge-fixed Lagrangian (28), it is not related to the Poisson structure (43) in a simple manner.
Note the non-Darboux form
$$\begin{align} \{C(\tau), \bar{C}(\tau^{\prime})\}_{DB}
~=~&\frac{e(\tau)^{-r}}{4\epsilon} \exp\left[\frac{R(\tau^{\prime},\tau)\chi^{\prime}}{2\epsilon}\right] {\rm sgn}(\tau\!-\!\tau^{\prime}) , \cr
\{ B (\tau), C(\tau^{\prime})\}_{DB}
~=~&\frac{r}{\epsilon} \frac{C(\tau)}{e(\tau)} \delta(\tau\!-\!\tau^{\prime}), \end{align} \tag{44}$$
to ensure that
$$ \{e(\tau)^r C(\tau), B(\tau^{\prime})\}_{DB}~=~0.\tag{45} $$
VII) Quantum BV formulation. Eqs. (30), (32) & (44) suggest that we should put $r=0$, so let us do this from now on. Inspired by the BFV-BRST transformations (34), we modify the BV Lagrangian (23) into
$$ \tilde{L}_{BV}~=~L_H +x^{\ast}_{\mu} g^{\mu\nu}(x)p_{\nu}C -\frac{1}{2}p_{\ast}^{\mu} \partial_{\mu}g^{\nu\lambda}(x)~p_{\nu}p_{\lambda} C +e^{\ast}\dot{C} + B \bar{C}^{\ast}. \tag{46} $$
One may show that the quantum master equation is now satisfied$^1$
$$ (\tilde{S}_{BV}, \tilde{S}_{BV})~=~0~=~\Delta\tilde{S}_{BV}. \tag{47} $$
The modification (46) does not alter the gauge-fixed Lagrangian (28) apart from putting $r=0$.
References:
David Tong, Lectures on String Theory, arXiv:0908.0333.
J. Polchinski, String Theory, Vol. 1, 1998; Section 4.2.
M. Henneaux and C. Teitelboim, Quantization of Gauge Systems, 1994; Chapter 17.
F. Bastianelli, Constrained hamiltonian systems and relativistic particles, 2017 lecture notes.
A. Cohen, G. Moore, P. Nelson & J. Polchinski, Nucl. Phys. B267 (1986) 143; Chapter 2.
C.M. Hull & J.-L. Vazquez-Bello, arXiv:hep-th/9308022; Chapter 2.
P. van Nieuwenhuizen, Lecture notes.
--
$^1$ We ignore boundary terms. Effectively this means that we impose pertinent boundary conditions, and limit gauge symmetry to the bulk.
$^2$ The $\epsilon$-dependence in the BFV action (37) comes only from the gauge-fixing fermion (38). The $\epsilon$-dependence can be removed via redefinition
$$ \epsilon B~\longrightarrow~ B, \qquad
\epsilon\bar{C}~\longrightarrow~ \bar{C}, \qquad
\frac{\chi}{\epsilon} ~\longrightarrow~ \chi, \qquad
\frac{\xi}{\epsilon^2} ~\longrightarrow~ \xi .\tag{48} $$
In the limit $\epsilon\to 0$, the infinities on the rhs. of the Poisson brackets (31) should be interpreted as zero, i.e. the corresponding canonical variables become decoupled.
