Faddeev-Popov Gauge Fixing Procedure

I want to know that does $F^{a}[A_{mu}] = 0$ condition used in Faddeev-Popov Quantization has unique solution $A_{mu}$ or is it $F^{a}[A^{theta}_{mu}] = 0$ should have unique $theta$ as solution where $A^{theta}_{mu}$ is gauge transformed $A_{mu}$ with parameter $theta^{a}(x)$ (but then won’t $theta = 0$ is also one of the solution). If this is the case that will mean that $F^{a}[A_{mu}] = 0$ initially does not exactly fixed the redundancy. I have this confusion because during the process we take
$1 = int Pi_{x, a}dtheta^{a}(x)delta(F^{a}[A^{theta}_{mu}])det[delta F^{a}[A^{theta}_{mu}(x)]/deltatheta^{b}(y)]$ (and prove that this is gauge invariant) and insert this in the path integral extracting out $int Pi_{x, a}dtheta^{a}(x)$ volume element and leaving path integral as
$int D[A]det[delta F^{a}[A^{theta}_{mu}(x)]/deltatheta^{b}(y)]delta(F^{a}[A^{theta}_{mu}])exp{iS}$ and now we again gauge transform with $-theta^{a}(x)$ parameter to remove dependence on $theta^{a}(x)$ from the path integral. Now my question is now while evaluating $det$ at $theta = 0$,
$det[delta F^{a}[A^{theta}_{mu}(x)]/deltatheta^{b}(y)]_{theta=0}$, can we also use $F^{a}[A_{mu}] = 0$ as one of the condition in evaluating this determinant (we cannot use this if our starting $A_{mu}$ does not satisfy gauge fixing condition $F^{a}[A^{theta}_{mu}] = 0$, assuming $theta$ is not zero).