Let $X\sim Exp(1)$ and $Y\sim Exp(\lambda)$, independent. Then, \begin{align} f_{X|Y=mX}(x) = \frac{f_{X,Y}(x,mx) }{\int f_{X,Y}(x,mx) \:dx }=\frac{f_X(x)f_Y(mx) }{\int f_X(x)f_Y(mx) \:dx } = \frac{e^{-(1+\lambda m)x}}{\int e^{-(1+\lambda m)x} dx} = (1+m\lambda)e^{-(1+\lambda m)x} \end{align} So $X|_{Y=mX} \sim Exp(1+m\lambda)$. That means $E[X|Y=mX]=\frac{1}{1+ \lambda m} < 1$ whenever $m>0$.
This makes sense mathematically. $X|_{Y=0}\sim Exp(1)$, $f_{X,Y}(0,0)$ is the same for all $m$, but $f_{X,Y}(x,mx)<f_{X,Y}(x,m'x)$ if $m'<m$.
The practical implication seems weird, though. Your friend gets to your house via one Poisson bus and one Poisson train. You expect someone to wait 1 minute for a train. But they tell you they waited $m$ times as long for the bus than they did for the train, and now you gotta revise your expectation about the train down?
Edit: I think the reason for the (seeming) violation of the Tower property is that I incorrectly defined $f_{X|Y=mX}$. It should instead be
$$ f_{X|Y=mX}(x) = \frac{xf_{X,Y}(x,mx) }{\int x f_{X,Y}(x,mx) \:dx} = \frac{xe^{-(1+\lambda m)x}}{\int xe^{-(1+\lambda m)x} dx} $$ Think about it like the flag of Seychelles: the width of the "ray" is twice as large if you go twice as far out. (Aside: this is indeed Borel's paradox) This means that \begin{align} E[X|Y=mX]= \frac{\int x^2e^{-(1+\lambda m)x} dx}{\int xe^{-(1+\lambda m)x} dx} = \frac{2}{1+\lambda m} \end{align} The distribution of slopes is the distribution of $M:=Y/X$ \begin{align} f_{M}(m) &= \int_0^\infty f_Y(y) f_{1/X}(m/y) y^{-1}\:dy\\ &= \int_0^\infty \lambda e^{-\lambda y}\left(\frac{y}{m}\right)^2 e^{-y/m} y^{-1} \:dy\\ &= \frac{\lambda}{m^2} \int_0^\infty y e^{-y(\lambda + \frac{1}{m}) } \:dy \\ &= \frac{\lambda}{m^2}\frac{1}{(\frac{1}{m} + \lambda)^2}\\ &= \frac{\lambda}{ (1+\lambda m)^2} \end{align} The Law of Iterated Expectaions holds for this definition of the conditional density: \begin{align} \int E[X|Y/X = m] \: dP(M\leq m) = \int_0^\infty \frac{2\lambda}{ (1+\lambda m)^3} \:dm = -\frac{1}{(1+\lambda m)^2}\bigg|_0^\infty = 1 = EX \end{align}