I am now studying QFT using Schwartz's book, and I am going through the part discussing about how the charge conjugation, parity, and time-reversal operator acting on various object should look like. He starts with how charge conjugation operator on the spinors act: $$C : \psi \rightarrow -i \gamma_{2} \psi^{*}$$
and then enlarges the action of $C$ on the other objects so that the Lorentz invariance is attained.
However, I find this logic to be somewhat strange. For example, suppose we have a Lagrangian $$\mathcal{L} = \phi^{*}(\Box + m^{2}) \phi + \frac{\lambda}{3!} \phi^{3}.$$ Then after imposing a transformation such that $\phi \rightarrow - \phi$, we see that $\mathcal{L}$ is not invariant under this. Then we just simply say that this Lagrangian does not have this symmetry and go on.
But the argument on the action of $C$, $P$ and $T$ doesn't go this way. We first declare that the Lagrangian we are interested in must be invariant under $C$, $P$ and $T$, and adjust the action of them on various objects to make that happen. For example, Schwartz says that the QED interaction term $eA_{\mu} \bar{\psi} \gamma^{\mu} \psi$ should be invariant under Lorentz transformations, and since $$C : \bar{\psi} \gamma^{\mu} \psi \rightarrow - \bar{\psi} \gamma^{\mu} \psi,$$ he declares that $C: A_{\mu} \rightarrow -A_{\mu}$. (Schwartz, Quantum Field Theory and the Standard Model, p.194~195))
Why can we do this for $C$, $P$ and $T$? Why do we not do this for others?
For example, for $$\mathcal{L} = \phi (\Box + m^{2} ) \phi + \frac{\lambda}{3!} \phi^{3},$$ and the transformation $\phi \rightarrow -\phi,$ why not we extend the transformation to $\lambda \rightarrow -\lambda$, so that the Lagrangian $\mathcal{L}$ is invariant under it?