The center of a group, G is defined as: Z = {a|ag = ga for all g ∈ G}.
Let $\phi$ be an automorphism of a finite group G to G. Show that for any a ∈ Z then $\phi(a) $∈ Z. Conclude that $\phi(Z) = Z$.
Proof: Let $\phi$ be an automorphism of a finite group G to G and let $a \in Z$. So $ag=ga$ for $g \in G$. What would I do next?
Continuing your proof: "...since $\;\phi\;$ is onto there exists $\;x\in G\;\;s.t.\;\;\phi(x)=g\;$ , and thus
$$\phi(a)g=\phi(a)\phi(x)=\phi(ax)=\phi(xa)=\phi(x)\phi(a)=g\phi(a)$$