G-equivariant automorphisms of G Let $G$ be group, and
$$
\operatorname{Aut}_G(G):=\{f\in\operatorname{Aut}G\mid f(xg)=f(x)g,\forall x,g\in G\}
$$
I want to show that in fact $\operatorname{Aut}_G(G)\cong G$.
$G\subset \operatorname{Aut}_G(G)$ is clear, since for any $g\in G$ we can define such a map $x\mapsto gx$, lies in $\operatorname{Aut}_G(G)$.
But how to prove the converse?
 A: So actually what you wanted to write is
$$Aut_G(G)=\{f:G\to G\ |\ f(xg)=f(x)g\text{ and }f\text{ is invertible}\}$$
i.e. the set of all $G$-equivariant automorphisms. Because $Aut(G)$ symbol (that you used inside) is reserved for group automorphisms. Note that we use the term "$G$-equivariant automorphism", but it doesn't mean that these are some special group automorphisms. Group automorphisms and $G$-equivariant automorphism are very different beasts. In fact a group automorphism is $G$-equivariant if and only if it is the identity. But your entire question only makes sense if we talk about $G$-equivariant automorphisms, not group automorphisms.
Note that $Aut_G(G)$ is a group together with function composition.

$G\subset \operatorname{Aut}_G(G)$ is clear

No, there is no such inclusion. There's an embedding, even isomorphism, but no inclusion: $G$ is never a subset of $Aut_G(G)$. It is important to be precise in mathematics.
We have a function
$$\Theta:G\to Aut_G(G)$$
$$\Theta(g)(x)=gx$$
We will show it is a group isomorphism.

*

*It is a group homomorphism: $$\Theta(gh)(x)=ghx=g\Theta(h)(x)=\big(\Theta(g)\circ \Theta(h)\big)(x)$$

*It is injective. We check this by evaluating the kernel. Note that $id:x\mapsto x$ is the neutral element of $Aut_G(G)$. Assume that $\Theta(g)=id$. Then $gx=x$ for all $x\in G$ and thus $g=e$ is the neutral element of $G$. Therefore $\ker\Theta$ is trivial and thus it is injective.

*It is surjective. Let $f:G\to G$ be a $G$-isomorphism. Then $f=\Theta\big(f(e)\big)$, because for any $x\in G$ we have $$\Theta\big(f(e)\big)(x)=f(e)x=f(ex)=f(x)$$
