Problem. Let $G$ be a finite group and $f\in \text{Aut}(G)$. Suppose $f$ maps more than three fourths of the elements of $G$ to their respective inverses. Prove that $f(x)=x^{-1}$ for each $x\in G$ (hence $G$ must be Abelian).
My attempt:
Define $A=\{ x\in G\mid f(x)=x^{-1}\}$. Then, by hypothesis, $|A|>\frac{3}{4}|G|$. Since $G$ is finite and $A\subset G$ all what's left to prove is that $|A|=|G|$ or $f(A)=G$... I don't know where to go from this...
Maybe I'm not going through the right path... Any help or hints are appreciated.
