= Solution
{c}
A <gauge-invariant> <path integral> counts all representatives of each <gauge orbit>. In a perturbative region with a regular <gauge fixing> slice, insert the <Faddeev-Popov gauge-orbit identity>
$$
1=\Delta_{\mathrm{FP}}[A]\int\mathcal D\omega\,\delta(F[A^\omega]-\sigma),\qquad
F[A]=\partial_\mu A^\mu+\lambda A_\mu A^\mu.
$$
The <Faddeev-Popov determinant> is the <functional Jacobian> converting integration over the <gauge transformation> parameter into integration over the condition. The redundant <gauge group> volume then factors out of normalized <correlation functions>.
Choose $A_\mu^\omega=A_\mu+\partial_\mu\omega$. Its infinitesimal variation gives
$$
\delta F=\partial^2\omega+2\lambda A^\mu\partial_\mu\omega,\qquad
\boxed{\mathcal M(x,y)=[\partial_x^2+2\lambda A^\mu(x)\partial_{x\mu}]\delta^{(4)}(x-y),\quad\Delta_{\mathrm{FP}}[A]=\det\mathcal M.}
$$
There is no derivative of $A$ in this expression: only $A_\mu$ is varied in $A^2$. For $\lambda=0$, the determinant is field independent and can be absorbed into normalization. For $\lambda\ne0$, it is a functional of the <gauge field>, so dropping it changes the <path integral>, even though the <gauge group> is Abelian. Constant residual <gauge transformations> are removed by boundary conditions or by dividing out their volume, as usual in perturbative <gauge fixing>.
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