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"?

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. 
