TITLE: A finite group such that every element is conjugate to its square is trivial
QUESTION [26 upvotes]: Suppose $G$ is a finite group such that $g$ is conjugate to $g^2$ for every $g\in G$.  
Here's a proof that $G$ is trivial.  First, observe that if $\lvert G\rvert$ is even, then $G$ contains an element $h$ of order $2$, in which case, $h$ is conjugate to $h^2=1$.  But this implies that $h=1$, so $h$ does not have order $2$.  By contradiction, $\lvert G\rvert$ is odd.  Then, by the Feit–Thompson theorem, $G$ is solvable.  In particular, this means that the derived series of $G$ terminates.  However, for any $g$ in $G$, there exists $a\in G$ such that $g^2=aga^{-1}$, i.e., $g=aga^{-1}g^{-1}\in G^{(1)}$.  It follows that $G^{(1)}=G$.  In fact, this shows that $G^{(n)}=G$ for all $n\geq 1$.  Since the derived series of $G$ terminates, this implies that $G$ must be trivial.  
While I'm convinced of the result, this proof is not particularly satisfying to me, since it relies on Feit-Thompson.  Is there an elementary proof that $G$ is trivial?

REPLY [12 votes]: Just to add to Lord Shark's answer, if you are curious, see Lemma 5.1 of the paper Gabriel Navarro, The McKay conjecture and Galois automorphisms, Annals of Mathematics, 160 (2004), 1129–1140.