Equivalences of skeletal categories are isomorphisms As explained here, equivalent skeletal categories are isomorphic. In my attempt to prove this fact, I started with skeletal categories $C$ and $D$ and equivalence functors $F: C \to D$ and $G:D \to C$ with $FG \simeq 1_D$ and $GF \simeq 1_C$. Via the natural isomorphisms, one obtains that both $FG$ and $GF$ are bijective on morphisms and fix objects (the former because each functor is fully faithful and the latter because for example $FGd \simeq d$ and so $FGd = d$, as $D$ is skeletal). My intention then was to see that both $GF$ and $FG$ are the corresponding identity functors, but in view of the aforementioned proof, it is not clear to me wether $F$ and $G$ are inverses, even though both functors are invertible. Does this hold?

It does not hold. There certainly exist pairs of mutually quasi-inverse equivalences between skeletal categories which are not strict inverses.

A good test case is that your categories are groups, which are always skeletal. An equivalence between groups $f:G\to H$ seen as categories is indeed just an isomorphism, but a map $g:H\to G$ is a quasi-inverse if and only if $gf$ and $fg$ are naturally isomorphic, that is, conjugate, to the appropriate identities. So if $G=H$ and $f=\mathrm{id}_G$, then $g$ can be any endomorphism of $G$ which is conjugate to the identity. More concretely, $g$ can be any inner automorphism.

So you can think of the indeterminacy in choice of your $G$ given $F$ as analogous to this. In short, there's no reason you should have $GF$ and $FG$ identities. You have to choose $G$ carefully, given $F$, to find the strict inverse.