= Solution
Choose the phase convention in the question and use the <gauge covariant derivative> $D_\mu=\partial_\mu+ieA_\mu$. For a charged <Dirac field>, the <quantum electrodynamics> Lagrangian is
$$
\mathcal L=-\frac14F_{\mu\nu}F^{\mu\nu}+\bar\psi(i\gamma^\mu D_\mu-m)\psi,
\qquad F_{\mu\nu}=\partial_\mu A_\nu-\partial_\nu A_\mu.
$$
The simultaneous local <gauge transformation> is
$$
\boxed{\psi'=e^{ie\chi}\psi,\qquad
\bar\psi'=\bar\psi e^{-ie\chi},\qquad A'_\mu=A_\mu-\partial_\mu\chi}.
$$
Direct differentiation gives $D'_\mu\psi'=e^{ie\chi}D_\mu\psi$: the derivative of the phase cancels the shift of $A_\mu$. The mass term and covariant kinetic term are therefore invariant. The <electromagnetic field tensor> is unchanged because mixed partial derivatives of $\chi$ commute. This proves classical <gauge invariance> of the coupled field action. A charged scalar similarly uses $(D_\mu\phi)^*D^\mu\phi$ and a potential depending only on $|\phi|^2$.
The electromagnetic coupling is $-eA_\mu\bar\psi\gamma^\mu\psi$. The matter equations give a conserved current $j^\mu=e\bar\psi\gamma^\mu\psi$, with $\partial_\mu j^\mu=0$. Constant $\chi$ gives the global symmetry and its conserved charge. When applying <Noether's theorem>, the sign of the parameterized Noether generator can be reversed without changing this conservation law; the transformation and coupling signs above remain fixed.
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