Homomorphism more than 3/4 the inverse

Question

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.

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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.

Answer 1

I think the point of this whole $3/4$ business is the following. If $G_1$ is the set of elements such that $f(x) = x^{-1}$, then if we look at left multiplication on $G$ by an element of $G_1$, more than half the elements have to make back into $G_1$.

Combining this with what we know about $f$ it should follow that any $g \in G_1$ commutes with more than $1/2$ the elements of $G$, so if you say Lagrange's theorem enough times it should follow that $G$ is abelian and $G_1$ generates $G$, which together imply the result.

Answer 2

(My girlfriend explained this to me.) After Anton's observation, it's sufficient to show that f = id if f fixes more than half of G. But the elements of G fixed by an automorphism form a group and this group has index less than 2 by assumption, hence is all of G.