Is it right to think of Parity as a change of basis in Dirac's Lagrangian? I'm trying to understand CPT symmetries in the Dirac Lagrangian but, so far, I've had more questions than answers. My naive view of CPT transformations is the following (please don't doubt to correct me):

After a parity transformation, a spinor $\psi$ is transformed into $\psi'$ with $\,\psi'(x,t)=\gamma^0\psi(-x,t)$. 
We usually omit the explicit position at which we're evaluating our fields (as we're lazy) and simply write $\,\psi'=\gamma^0\psi$. I can note then, that $\,\gamma^0$ is the matrix representation of the parity operator and I do tend to think of it as a change of basis. In addition, I also know that operators/matrices transform under change of bases as $\,M'=UMU^{-1}$.
If I wanted to know, let's say, how a gamma matrix transform under parity, I'd take $M=\gamma^\mu$ and $U=\gamma^0$.
For $\mu=1,2,3,4$,  $\,(\gamma^\mu)'=\gamma^0\gamma^\mu\gamma^0=(\gamma^\mu)^\dagger$ 
and for $\mu=5$, $\,(\gamma^5)'=\gamma^0\gamma^5\gamma^0=-\gamma^0\gamma^0\gamma^5=-\gamma^5$ 
These might not be true in reality, and the problems are evident when we compare these results with the literature:
(i) The second result implies that the pseudo-vector covariant bilinear $\bar{\psi}\gamma^5\psi$ is even under P (not odd).
(ii) The first one allows us to prove that the vector covariant bilinear $\bar{\psi}\gamma^\mu\psi$ transforms as it should (as a vector under P: odd in space, even in time) but it also implies that the Dirac Lagrangian is not parity invariant. Let's check it:
Let $\;\Delta_{(0)}\equiv-1$ and $\;\Delta_{(i)}\equiv1$
$\,\bar{\psi}'(i\gamma^{\mu'}\partial_{\mu'}-m)\psi'=\bar{\psi}\gamma^0\left(i(\gamma^{\mu})^\dagger(\Delta_{(\mu)}\partial_{\mu})-m\right)\gamma^0\psi=\bar{\psi}(i\gamma^{\mu}\Delta_{(\mu)}\partial_{\mu}-m)\psi\neq\bar{\psi}(i\gamma^{\mu}\partial_{\mu}-m)\psi$.
This is bad, but... isn't $M'=UMU^{-1}$ the right way to transform matrices?
I'm truly confused and also eager to read you feedback.
 A: 1) The matrix implementing parity will act as $P\cdot\psi(t,x)P^{-1}=\gamma^0\psi(t,-x)\equiv\psi_{ref}(t,x)$. $P$ is the parity matrix in question, not $\gamma^0$. Note that the operator acts as a unitary transformation, not as a matrix multiplying something like a wavefunction. 
2) It is clearer to see what parity does on states, as opposed to the fields. This is because traditionally, parity's action on vectors (such as $p$, the momentum) is well understood but on spinors (like $\psi$), its not. So, we want under parity, $$|p\rangle\to|-p\rangle\implies Pa_pP=a_{-p}$$
where I have used $P^2=1$ while implementing the unitary transformation on the creation operator $a_p$. See Peskin&Schroeder for more details.
3) Another way to proceed is to note that under parity, the 2 component weyl spinors undergo $$\psi_L\leftrightarrow\psi_R$$
roughly because the boost part generators of the $(1/2,0)$ and $(0,1/2)$ representations flips under parity as it sends $\vec{v}\to -\vec{v}$. You can use this split to see how bilinears transform under parity. For example, $$\bar{\psi}\gamma^5\psi=\psi^\dagger_L\psi_R-\psi^\dagger_R\psi_L$$
and it's clear that under parity, since $L\leftrightarrow R$, this  behaves as a pseudoscalar.
4) Since $\gamma^0$ is not the parity matrix, everything else you said about transformations of matrices under parity is incorrect. The gamma matrices are fixed once you fix a basis. There is not much meaning to transforming these matrices under parity and calling it a basis change.
