Suppose $G$ is a finite group and $f$ is an automorphism of $G$. If $f(x)=x^{-1}$ for more than $\frac{3}{4}$ of the elements of $G$, does it follow that $f(x)=x^{-1}$ for all $x$ in $G\ ?$

I know the answer is "yes," but I don't know how to prove it.

Here is a nice solution posted by administrator, expanded a bit:

Let $S = \{ x \in G: f(x) = x^{-1}\}$.

Claim:For $x$ in $S$, $S\cap x^{-1}S$ is a subset of $C(x)$, the centralizer of $x$.

*Proof:* For such $y$, $f(y) = y^{-1}$ and $f(xy) = (xy)^{-1}$. Now
$$x^{-1} y^{-1} = f(x)f(y) = f(xy) = (xy)^{-1} = y^{-1}x^{-1}.$$
So $x$ and $y$ commute.

Since $S\cap x^{-1}S$ is more than half of $G$, so is $C(x)$. So by Lagrange's Theorem, $C(x) = G$, and $x$ is in the center of $G$. Thus $S$ is a subset of the center, and it is more than half of $G$. So the center must be all of $G$, that is $G$ is commutative. Once $G$ is commutative the problem is easy.