Prove $f(x)=a*x*a^{-1}$ Group Homomorphism If $f:(G,*) \to (G,*)$ , $a$ in $G$
Prove that $f$ is a group homomorphism , where 
$$f(x)=a*x*a^{-1},\ \ \ \  x\in  G.$$

My answer:
We should to prove this :
$$f(x*y)=f(x)*f(y).$$
And from definition in the question:
$$\tag{1}f(x*y)=a*x*y*a^{-1}$$
and 
$$\tag{2}f(x)=a*x*a^{-1}$$
and
$$\tag{3}f(y)=a*y*a^{-1}.$$
But How can I prove $(1)=(2)*(3)$?

Sometimes, when you can't complete one side of an equality, you should try with the other one:
$$
f(x)*f(y)=(a*x*a^{-1})(a* y* a^{-1})=a*x*(a^{-1}*a)*y*a^{-1}\\
=a*x*e*y*a^{-1}
=a*x*y*a^{-1}
=f(x*y).
$$
