Simple proof of "$a$ and $a^{-1}$ have the same number of conjugates" I recently had to give a proof of this, I gave a correct proof but I feel that it was overly complicated, so the question here is "Find a simpler proof (one that belongs in "the book")".

Question: "Given a group $G$ and $a\in G$, prove that $a$ and $a^{-1}$ have the same number of distinct conjugates."
My answer (just a sketch):
Take $Cl(a)$ (conjugacy class of a) and $Cl(a^{-1})$, show that $x\in Cl(a) \iff x^{-1}\in Cl(a^{-1})$ (call this Statement 1).
Form the function $g$ on a set $S\subset G$, $g(S)=\{s^{-1}|s\in S\}$. Show that this is an involution, and thus a bijection.
Show that $Cl(a^{-1})=g(Cl(a))$ (by proving both $Cl(a^{-1})\subset g(Cl(a))$ and $g(Cl(a))\subset Cl(a^{-1})$ , by using Statement 1).
By definition of "same cardinality" ($\equiv$ there exists a bijection between them), conclude $Cl(a^{-1})=g(Cl(a)) \implies |Cl(a^{-1})|=|Cl(a)|$.

This took about 2.5 pages to fully explicate. Surely there must be a more concise proof for such a simple question?

Let $G$ act on itself by conjugation. The stabiliser of $a$ is $C(a)$, the centraliser of $a$, and $C(a) = C(a^{-1})$, so the number of conjugates of $a^{-1}$ is $(G:C(a^{-1})) = (G:C(a))$, which is the number of conjugates of $a$.
