Prove the map $\Phi:G\to\operatorname{Aut}G$ with $g\mapsto (x\mapsto g^{-1}xg)$ is an homomorphism. According to this article, 

A group homomorphism is a map $f:G \rightarrow H$ between two groups
  such that the operation is preserved: $f(g_{1}g_{2}) = f(g_{1})f(g_{2})$ for all $g_{1}, g_{2} \in G$, where the product on the left-hand side is in $G$ and on the right-hand side is $H$.

Question
Let G be a group and let $Aut(G)$ be the group of automorphisms of G.
(a) For any $g \in G$ define $\phi_{g} : G \rightarrow G$ by $\phi_{g}(x)=g^{-1}xg$.  Check that $\phi_{g}(x)$ is an automorphism.
(b) Consider the map
$\Phi : G \rightarrow Aut(G)$
$g \mapsto \phi_{g}$
Check that $\Phi$ is a homomorphism.
My Attempt
I did the first part.
Now, I need to do the second part of this problem.  Here is my attempt for the second problem.
Observe that if we evaluate $\Phi$ at $gh$ with arbitrary $x$, then:
$\Phi(gh) = \phi_{gh}(x) = (gh)^{-1}x(gh) = h^{-1}(g^{-1}xg)h = h^{-1}\phi_{g}(x)h = \phi_{h}(\phi_{g}(x))$.
The problem is that I'm stuck with this part.  I think that the operation of that $\Phi$ would be the composition of two functions.
Another thought: It would not make sense to say that:
$\Phi(gh) = \Phi(g)\Phi(h)$
because,
$\Phi (gh) = h^{-1}g^{-1}xgh$
But
$\Phi(g) = g^{-1}xg$
$\Phi(h) = h^{-1}xg$
$\Phi(g)\Phi(h) = g^{-1}xgh^{-1}xh$
The problem is that G is not assumed to be abelian (Well, if it is, then the equality holds.).  Then, we can't take the product of $\Phi$'s to match with $\Phi_{gh}$.  If I'm right, the solution to the second part of this problem is unlike the one to the first part of this problem.
Any comments or suggestions?
 A: What you have is an antihomomorphism, as you have shown. This is because you're using conjugation by $g^{-1}$. 
You should actually be looking at the function $\Phi$ that maps an element $g\in G$ to a function, namely that who sends $x$ to $gxg^{-1}$. So we can write it as $g\mapsto \phi_g$ where $\phi_g(x)=gxg^{-1}$.
The question is whether this is an homomorphism. So take $h,g\in G$. Then $\Phi(hg)$ is a function that maps $x$ to $$(hg) x(hg)^{-1}=h(gxg^{-1})h^{-1}$$ But this can be seen to be the composition of the function $\phi_h$ that sends $x\mapsto hxh^{-1}$ with that which sends $x\mapsto gxg^{-1}$, because $$\phi_h\circ \phi_g(x)=h(g xg^{-1})h^{-1}$$
It follows that $$\Phi(hg)=\phi_{hg}=\phi_h\circ \phi_g=\Phi(h)\circ \Phi(g)$$
Remember that our group operation in $\rm Aut $ is composition of functions.
A: Firstly I think you are confusing the function $\Phi$ with $\phi$. We have $\phi_g: x \mapsto g^{-1}xg$, but $\Phi: g \mapsto \phi_g$.
But actually using the function $\phi$ as defined, I think we get an antihomomorphism so that $\Phi(gh) = \Phi(h)\Phi(g)$. Using $\phi_g = gxg^{-1}$ would give the homomorphism.
A: Put $\Phi:g\longmapsto \phi_{g^{-1}}$
A: I think that the main problem is the way the internal operation in $\operatorname{Aut}(G)$ is define.
Let $*:\operatorname{Aut}(G) \times \operatorname{Aut}(G) \to \operatorname{Aut}(G)$  be an internal operation in $\operatorname{Aut}(G)$, where $f*g:=g \circ f$. It's easy to show that $(\operatorname{Aut}(G), *)$ is a group.
Now, if we do the same reasoning, we get:
$$
\Phi(gh) = \phi_{gh}(x) = (gh)^{-1}x(gh) = h^{-1}(g^{-1}xg)h = h^{-1}\phi_{g}(x)h = \phi_{h}(\phi_{g}(x))=(\phi_{h}\circ\phi_{g})(x)
$$
But then, we have that $(\phi_{h}\circ\phi_{g})(x)=\phi_{g}*\phi_{h}=\Phi(g)*\Phi(h)$. Which means that $\Phi$ is a group homomorphism.
