Starting with your penultimate line, we may proceed in a way similar to my answer to this question, by first noting that the components of the current are complex numbers and thus not matrix valued, such that
\begin{equation} (\psi^\dagger \gamma^0 \gamma^\mu \psi)^* = (\psi^\dagger \gamma^0 \gamma^\mu \psi)^\dagger. \end{equation}
We then use 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^0 \gamma^\mu \psi)^\dagger &= \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.