Does the $U(1)$ vector current flip under charge conjugation? The conserved $U(1)$ current of the Dirac Lagrangian is given by $j^\mu = \bar{\psi} \gamma^\mu \psi$, where $\bar{\psi} = \psi^\dagger \gamma^0$. As this is interpreted as electric current I would expect it to flip sign under charge conjugation. Charge conjugation Of a spinor $\psi$ is defined as $\psi^c = C\psi^*$ where $C$ is the unitary charge conjugation matrix that satisfies $C^\dagger \gamma^\mu C = -(\gamma^\mu)^*$ for all gamma matrices.
If I calculate the $U(1)$ current under charge conjugation I find
$$ j^\mu_c = \bar{\psi^c}\gamma^\mu \psi^c \\ = (C \psi^*)^\dagger \gamma^0 \gamma^\mu C \psi^* \\
= (\psi^\dagger)^* C^\dagger \gamma^0 C C^\dagger \gamma^\mu C  \psi^* \\
= (\psi^\dagger)^* (\gamma^0)^* (\gamma^\mu)^* \psi^*
\\
= (\bar{\psi} \gamma^\mu \psi)^*\\ 
= (j^\mu)^*  $$
Which hasn’t flipped sign as I thought it would. Have I made an error in my analysis?
Any hints would be appreciated. Thanks!
 A: For any fermion bilinear we have
$$
\psi^T_\alpha A_{\alpha\beta} \chi_\beta = - \chi^T_\beta A^T_{\beta\alpha}\psi_\alpha\,.
$$
So
$$
\begin{aligned}
(\bar\psi \gamma^\mu \psi)^* &= -\psi^* (\gamma^\mu)^\dagger(\gamma^0)^\dagger\psi
\\&= -\psi^* \gamma^0\gamma^0(\gamma^\mu)^\dagger\gamma^0\psi \\&=
-\bar\psi \gamma^\mu\psi\,.
\end{aligned}
$$
Where I used $(\gamma^0)^2 = 1$ and $\gamma^0(\gamma^\mu)^\dagger\gamma^0 = \gamma^\mu$. In the first line I applied the identity at the beginning with $\psi^T \to \bar\psi^*$ and $\chi \to \psi^*$.
A: Starting with your third to last line, we begin by rewriting
\begin{equation}
\begin{split}
(\psi^\dagger)^*(\gamma^0)^* (\gamma^\mu)^* \psi^* &= \psi^T \big[(\gamma^0)^\dagger\big]^T \big[(\gamma^\mu)^\dagger\big]^T (\psi^\dagger)^T \\
&= \big[\psi^\dagger (\gamma^\mu)^\dagger (\gamma^0)^\dagger \psi \big]^T\\
&= \psi^\dagger (\gamma^\mu)^\dagger (\gamma^0)^\dagger \psi
\end{split}
\end{equation}
where in going from penultimate to last line we have used that the components of the current are complex numbers and thus not matrix valued, such that we may drop the transpose. We may then proceed in a way similar to my answer to this question, using the following properties of the gamma matrices
\begin{align}
(\gamma^0)^\dagger &= \gamma^0, \\
(\gamma^\mu)^\dagger &= \gamma^0 \gamma^\mu \gamma^0, \\
(\gamma^0)^2 &= \mathbb{I}_{4},
\end{align}
where $\mathbb{I}_{4}$ is the identity to write
\begin{equation}
\begin{split}
\psi^\dagger(\gamma^\mu)^\dagger(\gamma^0)^\dagger \psi &= \bar{\psi}\gamma^\mu(\gamma^0)^2\psi\\
&= \bar{\psi} \gamma^\mu \psi.
\end{split}
\end{equation}
This is then the result $j^\mu_c = j^\mu$. This is a consequence of the charge conjugation symmetry of quantum electrodynamics. 
