Equivalent definitions of group action, isomorphism of GSet categories

Question

Context:
For group actions on a set, there are two habitual definitions and a bijection between objects from each definition:

1. A function $G\times A \rightarrow A$ with the usual properties
2. A group morphism $G\rightarrow Aut_{set}(A)=S_{A}$

We can then define two categories $GSet$ of group actions where objects respectively follow the precedent definitions. Morphisms, which correspond to equivariant functions, are defined in each category as respecting the following commutative diagrams:

1. For the first definition:$$\require{AMScd} \begin{CD} G\times A @>{\rho_A}>>A\\ @VVV \circlearrowleft @VVV \\ G\times A' @>{\rho_{A'}}>> A' \end{CD}$$
2. and for the second one: $\color{red}{\textbf{edit: }\text{as noted by Jules Essuie-Glace and Jendrik Stelzner, this definition is wrong.}}$
   $$\require{AMScd} \begin{CD} S_A @>{\phi}>>S_{A'}\\ @AAA \circlearrowleft @AAA \\ G @= G \end{CD}$$

Questions:
Intuitively, there should be an isomorphism between the two categories. The object part of such a functor is clear but I can't build the morphism part. Does such an isomorphism exist? Is more categorical formalism (such as viewing group actions as functors) necessary to have a more elegant formulation?

My attempt:
Informally, we have this diagram: $$\require{AMScd} \begin{CD} \{\text{1st-definition actions}\} @>\text{some 1st-Def equivariant morphism}>>\{\text{1st-definition actions}\}\\ @| @| \\ \{\text{2nd-definition actions}\} @>\text{some 2st-Def equivariant morphism}>> \{\text{2nd-definition actions}\} \end{CD}$$ Therefore, by denoting $\sigma$ the object bijection between 1st and 2nd definition actions, if $r$ and $s$ are the two equivariant morphisms, we should have: $$r = \phi \circ s \circ \phi^{-1}$$ I don't exactly know what types of objects to look for the following diagram to make sense. I've tried going to sets for one direction, but I keep getting mixed up in formalism and losing track of what type of objects I'm dealing with. For example, from a morphism $S_A\xrightarrow{\phi}S_{A'}$, we have by definition $\phi(id_A) = id_{A'}$. As sets, $id_A$ and $id_{A'}$ are the diagonals $A \mathrel{\tilde{=}}\Delta_A \subseteq A\times A$ and $A'\mathrel{\tilde{=}}\Delta_{A'} \subseteq A'\times A'$ and we obtain from $\phi$ a function $\hat{\phi}:A \rightarrow A'$. I am not able to show anything about it as I am not sure how to transfer equivariance of $\phi$ into the second category. Surely some language problem is happening. Similar problems arise for me in the second direction.

Answer 1

As noted by the gentlemen in the comments, the second definition for morphisms is wrong. In fact, the morphisms should be maps $f:A \rightarrow A'$ such that, with $\sigma_A :G \rightarrow S_A$ and $\sigma_{A'}:G\rightarrow S_{A'}$ the action morphisms and denoting $\sigma_g(\cdot)=\sigma_A(g)(\cdot)$, $\sigma'_g(\cdot)=\sigma_{A'}(g)(\cdot)$, we have: $$\forall g \in G, f\circ \sigma_g(\cdot)=\sigma'_g \circ f(\cdot)$$ which corresponds, with abuse of notation $g\cdot f = \sigma'_g(f)$ and $f(g\cdot)=f(\sigma_g(\cdot))$, to the commutative diagram: $$\require{AMScd} \begin{CD} A @>f>> A'\\ @V g\cdot V V \circlearrowleft @VV g\cdot V\\ A @>>f> A' \end{CD}$$ Similarly, the equivariant condition for the 1st definition means that, with the same notations as the initial post, we have: $$\rho_{A'}(g,f(\cdot))=f(\rho_A(g, \cdot))$$ We have already established that we have a bijection between 1st and 2nd definition objects. Now, define $\mathscr{F},\mathscr{G}$ that respectively send objects from the 2nd definition category to the 1st and vice versa. It is immediate that they are bijective on objects and therefore we have preimages for both cases which will be denoted with the pairs $\sigma_A -\rho_A$. For morphisms, we define $\mathscr{F}(f)=\mathscr{G}(f)=f$. In one direction ,we have: $$\rho_{A'}( \mathscr{F}(f))=\mathscr{F}(\sigma_{A'}) \circ \mathscr{F}(f)=\mathscr{F}(\sigma_{A'} \circ f)=\mathscr{F}(f\circ\sigma_A)=\mathscr{F}(f)\circ \mathscr{F}(\sigma_A) = \mathscr{F}(f) \circ \rho_A$$ Hence $\mathscr{F}(f)$ is indeed equivariant. The opposite direction follows the exact same proof. Skipping some detail, we see that $\mathscr{F},\mathscr{G}$ are indeed functors that preserve equivariant maps from one definition to the other and that they are inverse to each other, providing us with the wanted isomorphism.