In 1 the question why in lattice gauge theories with gauge group $G$, there was no need for gauge fixing to obtain finite path integrals was answered. Thus observables could be calculated as \begin{equation} \langle O \rangle = \frac{1}{Z} \int D[U] e^{-S[U]} O(U) \end{equation} when considering only the pure gauge case. However in one of the answer in 1 it was claimed that one could also fix the gauge and do the path integral over one specific slice, i.e. not over the space $G \times G \dots G$ but rather $G \times G \dots G / \sim$, where $\sim$ is the equivalence relation, collecting the field configurations of links $U_\mu(n) \in G$ in one equivalence class that only differ by a gauge transformation $U_{\mu}(n) \mapsto \Omega(n)U_{\mu}(n)\Omega(n+\mu)^\dagger$.
This would mean one could in a specific gauge do the path integral by only integrating over representatives of the gauge orbits. Now I could accept this fact, if every orbit had the same volume with respect to the haar measure. Otherwise one would weigh those (physical) configurations higher that happen to have a "larger" orbit under gauge transformation. This would be in contrast to fixing a gauge and then integrating, which would yield in weighing each physically distinct configuration "only once".
For example if the gauge group was finite, suppose $G= G_1 \cup G_2$ and $|G|<\infty$, where $G_i$ are the respective gauge orbits. For gauge invariant functions $f$ \begin{equation} \langle f \rangle = \frac{1}{|G|}\sum_{g \in G} f(g) = \frac{|G_1|}{|G|}f_1 + \frac{|G_2|}{|G|}f_2 \end{equation} But if I were to fix a gauge it would be \begin{equation} \langle f \rangle = \frac{1}{|G/\sim|}\sum_{g \in G/\sim} f(g) =\frac{1}{2}(f_1 + f_2) \end{equation} Thus even in this simple case, there would a difference in the calculation, if $|G_1| \neq |G_2|$.
It feels like saying: the theory is (by construction) already regulated so it is good enough and we leave it at that and do not worry about the fact that potentially fixing the gauge yields different outcomes.
