= Solution
Lorentz invariance requires $A^\mu$ and $J^\mu$ to transform as four-vectors, $\chi$ as a <scalar>, $m$ and $\mu_0$ as <scalars>, and the integration domain and <boundary conditions> to respect the transformation. Then every contracted term and $d^4x$ are Lorentz invariant.
Under $A_\mu\mapsto A_\mu+\partial_\mu\chi$, the field strength is unchanged. The source term changes by a boundary term plus
$$
-\int\chi\,\partial_\mu J^\mu,d^4x,
$$
which vanishes for a conserved current and suitable boundary behaviour of $\chi$. The mass term is not gauge invariant. Thus the action has the usual local gauge invariance when
$$
\boxed{m=0,\qquad \partial_\mu J^\mu=0,}
$$
with the gauge <function> or its boundary contribution suitably controlled. For $m>0$, the mass term breaks this gauge symmetry.
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