= Solution
Use the printed <gamma matrix> conventions throughout, in particular $\gamma^5=-i\gamma^0\gamma^1\gamma^2\gamma^3$. The transpose identity for $\gamma^0$ gives $C\gamma^0=-\gamma^0C$, and its <Hermitian conjugate> gives the corresponding identity for $C^\dagger$. Thus the <Dirac adjoint> of the charge-conjugated field is
$$
\bar\psi^{\,C}=\eta_C^*\psi^T\gamma^0C^\dagger\gamma^0=-\eta_C^*\psi^TC^\dagger.
$$
Reordering the anticommuting components of the <fermion bilinear> supplies a minus sign:
$$
\bar\psi^{\,C}\psi^C=-\psi^TC^\dagger C\bar\psi^T
=\bar\psi(C^\dagger C)^T\psi.
$$
Invariance for arbitrary fields therefore requires \b[$C^\dagger C=I$], so the charge-conjugation matrix is a <unitary matrix>. This is the required normalization constraint; it is not a bosonic reordering calculation.
For a spinor matrix $\Gamma$, the same calculation and the given $C^{-1}=C^T$ give
$$
(\bar\psi\Gamma\psi)^C=\bar\psi(C^{-1}\Gamma C)^T\psi
=\bar\psi C^{-1}\Gamma^TC\psi.
$$
With $V^\mu=\bar\psi\gamma^\mu\psi$ and $A^\mu=\bar\psi\gamma^\mu\gamma^5\psi$, the supplied transpose identities imply
$$
C^{-1}\gamma^{\mu T}C=-\gamma^\mu,\qquad
C^{-1}(\gamma^\mu\gamma^5)^TC
=-\gamma^5\gamma^\mu=\gamma^\mu\gamma^5.
$$
Consequently \b[the <vector current> is C-odd and the <axial current> is C-even]: $V^\mu\mapsto-V^\mu$ and $A^\mu\mapsto A^\mu$.
For <parity>, put $\Lambda=\operatorname{diag}(1,-1,-1,-1)$. The <gamma matrix> identities $\gamma^0\gamma^\mu\gamma^0=\Lambda^\mu{}_{\nu}\gamma^\nu$ and $\gamma^0\gamma^5\gamma^0=-\gamma^5$ give
$$
\boxed{V^\mu(x)\stackrel P\longmapsto\Lambda^\mu{}_{\nu}V^\nu(x_P),\qquad
A^\mu(x)\stackrel P\longmapsto-\Lambda^\mu{}_{\nu}A^\nu(x_P).}
$$
Thus $V^0$ is P-even and $\mathbf V$ P-odd, whereas $A^0$ is P-odd and $\mathbf A$ P-even. The <axial current> transforms as a <pseudovector>.
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