The conditions are the following.
Let $(G,\circ)$ be a group, and $a$ an element of $G$.  

Prove that this is a group isomorphism:

$$\phi : G \to G : \quad\phi(x) = a^{-1}\circ x\circ a$$

I know that I have to prove it is:

1. well-defined
2. bijective
3. homomorphic

The main question I have is can I switch the e.g. $a$ and $x$? So the function would be $\phi(x)= a^{-1} \circ a \circ x$, then it would be pretty easy, but since $G$ is not abelian I am not quite sure how to proof the 3 axioms for the isomorphism.

Can anyone help me?