Reconstructions of Groups From Category of $G-\mathbf{Sets}$; Construction of a Group Homomorphism I try to come up with a proof of the following statement, but I find it a little difficult. I hope that I can get some help from someone on this site. I think this is what they give a proof of, on Ncatlab - Tannakian Duality (at the section $G-\mathbf{Sets}$). But I can't really follow that proof: https://ncatlab.org/nlab/show/Tannaka+duality#ForPermutationRepresentations.
Statement. Let $F:G-\mathbf{Sets}\to\mathbf{Sets}$ be the forgetful functor, where $G-\mathbf{Sets}$ is the category of sets equipped with a group action by the group $G$. I'm trying to understand the proof of the following fact
$$\operatorname{Aut}(F)\cong G.$$

What I have done
I have managed to construct a map
$$\varphi:G\to\operatorname{Aut}(F)$$
This was done by the following rule $\varphi(g)=\eta^g$, where $\eta_S^g:S\to S$ is defined by $\eta_S^g(s)=s\cdot g$. It is straightforward to check that this gives a natural transformation from $F$ to $F$ and that it also is a group homomorphism.
However, the other way is more problematic for me. I want to find a map
$$\psi:\operatorname{Aut}(F)\to G.$$
That is, given a natural transformation $\eta$, I want to assign it to a group element $g\in G$.
The natural transformation $\eta$ is defined by the following commutative diagram
$\require{AMScd}$
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
F(X) & \ra{\eta_X} & F(X) \\
\da{F(f)} & & \da{F(f)} \\
F(Y) & \ra{\eta_Y} & F(Y) & \\
\end{array}
$$
where $\eta_X$ is a morphism in $\mathbf{Sets}$ and $f:X \to Y$ is a morphism in the category $G-\mathbf{Sets}$. Since $F$ is just the forgetful functor, the above diagram reduces to
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
X & \ra{\eta_X} & X \\
\da{f} & & \da{f} \\
Y & \ra{\eta_Y} & Y & \\
\end{array}
$$
Concerns and Questions
In the definition of natural transformation - I have that - given any $G-\text{Set}$ $X$, $\eta_X:F(X)\to F(X)$ is a morphism. A natural $G-\text{Set}$ is simply to take $X=G$ and to let it act on itself through the group structure: $$\varphi: G\times G\to G \\ (g,s)\mapsto g\cdot s.$$
So the commutative diagram now becomes
$$
\newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!}
\newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.}
%
\begin{array}{llllllllllll}
G & \ra{\eta_G} & G \\
\da{f} & & \da{f} \\
Y & \ra{\eta_Y} & Y & \\
\end{array}
$$
Remark 1. I remember a professor told me that the morphism $\eta_G$ is totally understood by what it does to the identity element $e\in G$ (from which I should be able to understand how to construct the group homomorphism),
$$e\mapsto \eta_G(e).$$
I don't really understand what the above means. I think I have misunderstood something about the forgetful functor. When I think about the forgetful functor $F:A\to B$, I think that the functor forgets everything that is present in $A$, but isn't present in $B$. In our case, it forgets the structure of group actions. And so, in particular, I cannot use the property of being a $G$-equivariant map. Only the properties of being a set-theoretic map.
Question 1.
If $\eta_G(e)=s$, and if I would like to make sense of what the professor told me, I think I would reason something as follows
$$\eta_G(g)=\eta_G(e\cdot g)=\eta_G(e)\eta_G(g)=s\eta_G(g).$$
where I in the second equality used the property of being a group homomorphism. But on the other hand, if I want to treat it as a group homomorphism, then I think I had to do it to start with. That is, $\eta_G$ must map identities to identities (in order to be consistent in my reasoning). So I think my argument fails.
My question is: What does he mean?
I don't think what I did above makes any sense. But I think I have seen others using the properties of the morphisms in the category $A$, after having applied the forgetful functor, hence my reasoning. Once again, I am not really sure what I am doing. So I may very well be wrong.
Question 2. How does this tell me where to map a natural transformation?
Given a $\eta\in\operatorname{Aut}(F)$, where do I map it? Do I map it as follows
$$\eta\mapsto \eta_G(e)?$$
Doing so, do I know that I have exhaustively told where to map every natural transformation?
Question 3. I guess I also, somehow, have to use the commutative diagram in the definition of the natural transformation when I construct the group homomorphism, which I haven't done? I guess my suggestion above is not the correct way to do it. Do you have any ideas how I can construct the map?
I would be really happy I could have any help from someone on this site to understand this better. Because I am really lost, and confused.
Best wishes,
Joel
 A: I am going to use left $G$-sets, not right.
Question 1 & 3.
You cannot write $\eta_G(e\cdot g)=\eta_G(e)\eta_G(g)$, we are not assuming $\eta_G:G\to G$ is a group homomorphism, only that it is a morphism of $G$-sets. You can ue that to say $\eta_G(g\cdot e)=g\cdot\eta_G(e)$ though (which you'd reverse the order of if you insist on right group actions).
Consider your commutative diagram again:
$$\require{AMScd}
\begin{CD}
G @>{\eta_G}>> G \\
@VVV @VVV \\
Y @>{\eta_Y}>> Y
\end{CD}$$
Here, we can let the map $G\to Y$ be the evaluation-at-$y$ map $g\mapsto gy$ where $y\in Y$ is fixed (note the evaluation map is also useful in establishing the orbit-stabilizer theorem - its fibers are cosets of $y$'s stabilizer). Then we chase the diagram starting from $e\in G$ in the top left.
If we follow the upper-right path, we get $e\mapsto \eta_G(e)\mapsto \eta_G(e)y$. In the lower-left path, $e\mapsto y\mapsto \eta_Y(y)$. Therefore we may equate $\eta_Y(g):=\eta_G(e)y$. That is, every automorphism $\eta$ applied to a $G$-set $Y$ is just applying a particular group element $\eta_G(e)\in G$.
Qusetion 3.
Yes, $\eta\mapsto \eta_G(e)$. This applies for all $\eta\in\mathrm{Aut}\,F$.
