Let G be a group and let G act on itself by conjugation, so each g in G maps G to G by $\displaystyle x\mapsto gxg^{-1}$. For each g in G, prove that conjugation by g is an isomorphism from G onto itself. Deduce that x and $\displaystyle gxg^{-1}$ have the same order for all x in G and that for any subset A of G, $\displaystyle |A|=|gAg^{-1}|$.

I have shown this an isomorphism. I am not sure how to show x and $\displaystyle gxg^{-1}$ have the same order.