Why is $\text{Aut}(F)$ of the forgetful functor $F$ on $G$-sets isomorphic to $G$?

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 $g$.

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 $$\text{Aut}_{\mathcal F}(F)=\text{Aut}_{\mathcal C'}(G)=G.$$

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.