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.

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.

I believe that you may have missed a minus sign in going from your third to fourth line of working, based on your definition $C^\dagger \gamma^\mu C = -(\gamma^\mu)^*$. With this correction, we would have

\begin{equation} \begin{split} (\psi^\dagger)^* C^\dagger \gamma^0 C C^\dagger \gamma^\mu C \psi^* &= -(\psi^\dagger)^* (\gamma^0)^* (\gamma^\mu)^* \psi^* \\ &= -(\psi^\dagger \gamma^0 \gamma^\mu \psi)^*. \end{split} \end{equation}

If this is true, then 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 $j^\mu_c = -j^\mu$ that you were expecting.

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.