Let $G$ a finite group.
Let $f: G \to G$ an automorphism such that at least half the elements of the group are sent to their inverses, i.e $$\mathrm{card}(\{g \in G|f(g) = g^{-1}\}) \geq \frac{\mathrm{card}(G)}{2}.$$
Question : Is $f$ an involution ?
(I know that people consider which the identity is not an involution, but for me an involution is only a function $h$ such that $h\circ h = \mathrm{id}$).
Thanks.
I think yes.
Let $X$ be the set of elements of $G$ inverted by $f$, and $H$ be the subgroup generated by $X$.
An element $h$ of $H$ is a product of elements inverted by $f$, so $H$ is invariant under the action of $f$ (that is $f(H)=H$), and if we apply $f$ two times on $h$ we obtain $h$. This amounts to saying that $f$ induces an involution on $H$.
If $H=G$ we are done. If $H <G$, then $1 < |G:H| \leq |G|/|X| \leq 2$, it follows that $|G:H|= |G|/|X|=2$, so that $H=X$ (since $H$ contains $X$). This shows that $H$ is normal, and as it is invariant under $f$, $f$ induces an automorphism on the cyclic group of order 2, $G/H$. It follows that $f$ is the identity on $G/H$, so if $y \in G-H$, then $f(y)=yh$ for some element $h\in H=X$. We have $$f^2(y)=f(yh)=f(y)h^{-1}=yhh^{-1}=y.$$ Now any element of $G$ which does not lie in $H$ has the form $yx$ for some $x \in H$. It follows that $f^2(yx)=f^2(y)f^2(x)=yx$. If we take an element $x$ of $H=X$, then clearly $f^2(x)=x$. Thus $f$ is an involution.