Functor $F: \mathcal{C} \to G_1$-Sets is an equivalence of categories given a natural isomorphism to equivalence $H: \mathcal{C} \to G_2$-Sets This is the argument I need in order to understand a proof of a certain theorem. I think that I'm mostly confused by the formalism, so I kindly ask for your help.
Suppose we have a functor $F: \mathcal{C} \to G_1\text{-Sets}$ for some group $G_1$ that is known to be an equivalence of categories. The goal is to prove that another functor $H: \mathcal{C} \to G_2\text{-Sets}$ is also an equivalence. The proof starts with giving a group isomorphism $\psi$ between $G_1$ and $G_2$. We also have a result for an object $q \in \mathcal{C}$ there is a bijection $\phi_q$ between sets $F(q)$ and $H(q)$ in the sense that we drop group actions and state that the underlying sets are bijective.
It is stated that if we manage to prove that the bijection $\phi_c$ is compatible with the group action, the result that $H$ is an equivalence of categories follows. That is, we need to prove that the diagram

commutes. I denote $F(q), H(q)$ to be plain sets as above, as opposed to them being $G\text{-Sets}$ like they are supposed to be. Group actions are denoted with $\rho$.
I don't understand why this is supposed to prove that $H$ is an equivalence of categories. Also note that a group can be thought of as a groupoid with a single object, group elements forming an automorphism group. Then $\rho$ can be thought of as a functor from this groupoid to the category of sets. The diagram is then seen as some sort of a natural transformation. I mean the if we reason about the G-Set as a functor, then the diagram above is equivalent to
,
where the diagram commutes for every automorphism $g_1$ and where $F(G_1), H(G_2)$ denote images of a single object of groupoid as described above. Both of these images also depend on $q \in \mathcal{C}$ as $F, H$ in the figure above are functor images of $q$.
I am really confused at this point since this is actually not a natural transformation. In this case we have that functors $F, H$  have different domains, one of them being a $G_1$-groupoid, the other being a $G_2$-groupoid. So this is, strictly speaking, not the natural transformation. At this point, I find myself lost in formalism. What exactly the commutativity of the diagram is supposed to show? I fail to understand why we would need this argument at all. I had some vague thoughts that it is supposed to show that functors $F,H$ are isomorphic via a natural transformation, and then it would somehow follow that $H$ is essentially surjective, and maybe it would also follow that $H$ is fully faithful as well. But I can't really think in this direction since functors of G-Sets I described first have different codomains, and functors from groupoids have different domains.
 A: Note that $\psi$ induces an isomorphism of categories $\psi_*:G_1\text{-Sets}\to G_2\text{-Sets}$ which just takes a $G_1$-set and considers it as a $G_2$-set by identifying $G_1$ and $G_2$ via $\psi$.  Your first diagram commuting then just says that for each $q$, $\phi_q$ is a morphism (and thus an isomorphism) of $G_2$-sets from $\psi_*(F(q))$ to $H(q)$: it preserves the action of $G_2$.
That is, the $\phi_q$'s can be considered as a transformation between the functors $\psi_*\circ F$ and $H$.  If you additionally knew that this was a natural transformation, then it would be a natural isomorphism.  Now since $F$ is an equivalence and $\psi_*$ is an isomorphism, $\psi_*\circ F$ is also an equivalence.  As you have guessed, then, it is possible to show that any functor that is naturally isomorphic to an equivalence of categories is an equivalence of categories, to conclude that $H$ is an equivalence.
However, all of this is dependent on knowing that $\phi_q$ is natural in $q$, and the diagram you are asking about has nothing to do with that naturality (it only has to do with $\phi_q$ being a morphism of $G_2$-sets).  So the information you have described is not enough to prove that $H$ is an equivalence.  You need to also show that $\phi_q$ is natural in $q$.  That is, you need to show that if $f:q\to r$ is a morphism in $\mathcal{C}$ then the composition of $F(f):F(q)\to F(r)$ and $\phi_r:F(r)\to H(r)$ is equal to the composition of $\phi_q:F(q)\to H(q)$ and $H(f):H(q)\to H(r)$.  (Note that here we are talking about a natural transformation between the functors $\psi_*\circ F$ and $H$, so every time I write $F$ I really mean $\psi_*\circ F$.  But it makes no difference here, since $F(q)$ and $F(r)$ have the same underlying sets as $\psi_*(F(q))$ and $\psi_*(F(r))$ and $\psi_*(F(f))$ is the same function as $F(f)$.)
