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.