Why is $\text{Aut}(F)$ of the forgetful functor $F$ on $G$-sets isomorphic to $G$? Here's something I've been trying to scratch out recently.

Let $G$ be a group and $\text{Set}(G)$ the category of $G$-sets. Let $F\colon\text{Set}(G)\to\text{Set}$ be the forgetful functor sending a $G$-set to its underlying set. Show that $\text{Aut}(F)$ is naturally isomorphic to $G$.


I've been trying to view $F$ as an object in the category of functors from $\text{Set}(G)\to\text{Set}$ where the morphisms are natural transformations. I think if $H$ is a natural transformation in $\text{Aut}(G)$, then for any $G$-set $X$, $H$ associates a morphism $H_X\colon F(X)\to F(X)$ such that for any morphism $f\colon X\to Y$ of $G$-sets, then $H_Y\circ F(f)=F(f)\circ H_X$?
I'm not really sure where I'm going with this. Given such an $H$, what is the natural $g\in G$ with which to associate it, and how would I get to this "natural isomorphism"?
 A: Define $u:\text{Aut}(F)\to G$ and $v:G\to\text{Aut}(F)$ by 
$$
u(a):=a_G(1),\quad v(g)_X(x):=gx.
$$
It suffices to show: (1) $u$ is a group morphism, (2) $u\circ v=\text{Id}_G$, (3) $v\circ u=\text{Id}_{\text{Aut}(F)}$. 
(1) We have $u(ab)=(ab)_G(1)=a_G(b_G(1))$ and $u(a)u(b)=a_G(1)b_G(1)$. Define $f:G\to G$ by $f(g):=gb_G(1)$. As $f$ is a $G$-map, it commutes with $a_G$, and we get (1) by evaluating $a_G\circ f=f\circ a_G$ on $1$. 
(2) We have $u(v(g))=v(g)_1=g$. 
(3) We have $v(u(a))_X(x)=u(a)(x)=a_G(1)(x)$. It should be equal to $a_X(x)$. 
Define $f:G\to X$ by $f(g):=gx$. Being a $G$-map, it satisfies 
$$
a_X\circ F(f)=F(f)\circ a_G,
$$ 
and it suffices to evaluate this equality on $1$. 
EDIT. Here is a selfcontained version of Zhen Lin's answer. 
Let $\mathcal C$ be a category, $\mathcal C'$ the opposite category, $\mathcal S$ the category of sets, and $\mathcal F$ the category whose objects are the functors from $\mathcal C$ to $\mathcal S$ and whose morphisms are the functorial morphisms.  
It is straightforward to check the following statements. 
The formula 
$$h(X):=\text{Hom}_{\mathcal C}(X,?)$$ 
defines a functorial morphism 
$$h:\mathcal C'\to \mathcal F.$$ 
Let $X$ be an object of $\mathcal C$. The formulas 
$$u(t):=t_X(\text{Id}_X),\quad v(a)_Y(f):=F(f)(a)$$ 
define functorial morphisms
$$u:\text{Hom}_{\mathcal F}(h(X),F)\to F(X),\quad v:F(X)\to \text{Hom}_{\mathcal F}(h(X),F)$$ 
which are functorial in $X$. Moreover 

$u$ and $v$ are inverse isomorphisms. 

In particular we have functorial isomorphisms 
$$\text{Hom}_{\mathcal F}(h(X),h(Y))=\text{Hom}_{\mathcal C}(Y,X)=\text{Hom}_{\mathcal C'}(X,Y),$$ 
$$\text{Aut}_{\mathcal F}(h(X))=\text{Aut}_{\mathcal C'}(X).$$ 
Now let $G$ be a group, $\mathcal C$ the category of $G$-sets, $F$ the forgetful functor. Then the formulas 
$$a_X(f):=f(1),\quad b_X(x)(g):=gx$$ 
define functorial morphisms
$$a:h(G)\to F,\quad b:F\to h(G).$$ 
Moreover $a$ and $b$ are inverse isomorphisms. This gives in particular canonical isomorphisms $$\text{Aut}_{\mathcal F}(F)=\text{Aut}_{\mathcal C'}(G)=G.$$ 
A: Everyone is working too hard. This follows from exactly two applications of the Yoneda lemma and nothing else. 
First, the forgetful functor is represented by $G$ as a $G$-set, so its automorphisms can be identified with the automorphisms of $G$ as a $G$-set by a first application of the Yoneda lemma.
Next, $G$ as a $G$-set is the unique representable $G$-set: that is, as a functor from the category $BG$ consisting of one object $\bullet$ with automorphism group $G$ to $\text{Set}$, it's the unique representable one, so its automorphisms can be identified with the automorphisms of $\bullet$ by a second application of the Yoneda lemma. But this is $G$ by definition. 
A: If $C$ is any algebraic category with forgetful functor $U : C \to \mathrm{Set}$ and left adjoint $F : \mathrm{Set} \to C$, then the free algebra on one generator $F(1)$ represents $U$. By the Yoneda lemma, it follows that $\mathrm{End}(U) \cong \mathrm{End}(F(1))$ as monoids. We also have a bijection $\mathrm{End}(F(1)) = \mathrm{Hom}_C(F(1),F(1)) \cong \mathrm{Hom}(1,U(F(1)) \cong U(F(1))$, which makes the underlying set of $F(1)$ a monoid such that this bijection becomes an isomorphism of monoids. Thus, we have $\mathrm{End}(U) \cong F(1)$ as monoids.
When $C=M-\mathrm{Set}$ for some monoid $M$, then $M$ is the free $M$-set on one generator and its monoid structure coincides with the given one.
When $C=R-\mathrm{Mod}$ for some ring $R$, then $R$ is the free $R$-module on one generator and its monoid structure is just the multiplicative structure.
When $C=R-\mathrm{CAlg}$ for some commutative ring $R$, the free commutative $R$-algebra on one generator is the polynomial algebra $R[T]$, and again the monoid structure is the multiplicative one.
You can add more categories as you like :).
A: This is a Yoneda-type argument. First, observe that $F \cong \textrm{Hom}_G(G, -)$, where $G$ is considered as a $G$-set by equipping it with the regular left action. So a natural transformation $F \Rightarrow F$ is also a natural transformation $\textrm{Hom}_G(G, -) \Rightarrow \textrm{Hom}_G(G, -)$, and by the Yoneda embedding, there is a natural bijection between such natural transformations and $\textrm{Hom}_G(G, G)$, which is just $G$ itself. (This implies all such natural transformations are in fact natural isomorphisms.)
More explicitly, let $\eta : F \Rightarrow F$ be a natural transformation, and let $g = \eta_G(e)$. Then $\eta_X(x) = g \cdot x$: indeed, if $f : G \to X$ be the $G$-equivariant map determined by $f(e) = x$, then $F(f) \circ \eta_G = \eta_X \circ F(f)$, so $\eta_X(x) = f(g) = g \cdot x$ (with some abuse of notation). Conversely, it is clear that this defines a unique natural transformation $\eta : F \Rightarrow F$ for each $g$.
