Consider a complex scalar particle $\phi$ coupled to an electromagnetic field. The Lagrangian is given by
$$ \mathcal{L} =(D_\mu \phi)^* D^\mu \phi - m^2 \phi^2 - \frac{1}{4} F_{\mu \nu} F^{\mu \nu}$$
where $D_\mu = \partial_\mu - ie A_\mu $ and $F_{\mu \nu} = \partial_\mu A_\nu - \partial_\nu A_\mu$. This Lagrangian has global $U(1)$ symmetry under $\phi \rightarrow e^{i \alpha} \phi$, $\phi^* \rightarrow e^{-i\alpha } \phi$. The corresponding Noether current is given by
$$ j^\mu(\phi) = -ie[\phi^* D^\mu \phi - (D^\mu \phi)^* \phi] = -ie(\phi^* \partial_\mu \phi -\phi \partial_\mu \phi^* - 2ieA_\mu|\phi|^2)$$$$ j^\mu(\phi) = -i[\phi^* D^\mu \phi - (D^\mu \phi)^* \phi] = -i(\phi^* \partial_\mu \phi -\phi \partial_\mu \phi^* - 2ieA_\mu|\phi|^2)$$
and is interpreted as the electric current, as discussed in this question
One would expect that under charge conjugation $\phi \rightarrow \phi^*$ that the electric current would change sign. If I replace $\phi $ with $\phi^*$ in the current above, I find
$$j^\mu(\phi^*)= -ie(\phi \partial_\mu \phi^*-\phi^* \partial_\mu \phi - 2ie A_\mu |\phi|^2) \neq -j^\mu(\phi) $$$$j^\mu(\phi^*)= -i(\phi \partial_\mu \phi^*-\phi^* \partial_\mu \phi - 2ie A_\mu |\phi|^2) \neq -j^\mu(\phi) $$
So it hasn't flipped sign due to the $|\phi|^2$ term. What is going on?