I'm looking for a derivation of the often quoted fact that the conservation of electric(!) current $j^{\mu} = (c \rho, \vec{j})$ in relativistic classical electrodynamics is an explicit consequence of Noether theorem. In other words that that the electric current $j^{\mu}$ is a Noether current with respect gauge transformation $A_{\mu} \to A_{\mu} + \delta A_{\mu}= A_{\mu} + \partial_{\mu} \chi$ where $\chi$ is any twice differentiable scalar function that depends on position and time.
Recall from classical Electrodynamics $S = \int_V \mathcal{L} dV dt$ with Lagrangian
$$\mathcal{L} := -\frac{1}{c} A_{\mu} j^{\mu} -\frac{1}{16 \pi}F^{\mu \nu} F_{\mu \nu}$$
with field tensor $F_{\mu \nu} = \partial_{\mu} A_{\nu}- \partial_{\nu} A_{\mu}$ and in context of variational calculus of action functional $S$ the conserved function, also called Noether current for action $\phi \to \phi + \delta \phi$ on Lagrangian $\mathcal{L}(\phi, \dot{\phi}, t)$ is defined by
$$J^{\mu} := \frac{\partial \mathcal{L}}{\partial \dot{\phi}} \delta \phi$$
Now I not see why in context of relativistic classical electrodynamics and gauge transformation of the $4$-potential $A_{\mu}$ (that is $\phi := A_{\mu}$ and $\delta \phi = \partial_{\mu} \chi$ the obtained Noether current $J^{\mu} $ coinsides exactly with classical electric current $j^{\mu} = (c \rho, \vec{j})$?
Does anybody know where I can find a derivation of that?
My attempts:
By equation of motion and definition Maxwell-Lagrangian we have $\frac{\partial F^{\mu \nu} F_{\mu \nu}}{\partial \dot{A_{\mu}}}= -4 F^{\mu \nu}$, so $\frac{\partial \mathcal{L}}{\partial \dot{A_{\mu}}}= \frac{1}{4 \pi}F^{\mu \nu}$ and $\delta A_{\mu}= \partial_{\mu} \chi$. Why does it imply $J^{\mu} = (c \rho, \vec{j})$. I not see it. Although I found in PSE some questions dealing which similar problem I nowhere found a source containing a full complete derivation of the claim, but on the other hand incountable many sources using this a fact.
