Finite groups with automorphism mapping $a/b$ of the elements of $G$ to their own inverses? Case $a/b=3/4$? I was helping a friend prepare for his intro abstract final and he mentioned the professor had once asked the question: name a group and an automorphism that takes $3/4$ of the elements of the group to their own inverses (for instance, the dihedral group $D_4$ of order $8$, with identity automorphism). I tried to figure out how to approach this question in general but can't see how.

1) Can we construct all such groups?

(it is asserted in comment to the answer below that these are precisely those finite groups whose center has index 4)

2) Given a rational number $a/b\in [0,1]$ does there exist a finite group $G$ and an automorphism $f$ such that $f$ maps exactly $a/b$ elements of $G$ to their own inverses?

($a/b=1$ is achieved precisely for the inversion map on an abelian group; otherwise $a/b\le 3/4$ according to the answer below)

3) Also, can these questions make sense in infinite groups?

 A: This may be a well-known chestnut? (well-known to those that know it well, that is)
The fraction can never be between 3/4 and 1. To prove this, suppose $\phi\colon G\to G$ is an automorphism of $G$ that sends more than 3/4 of the elements of $G$ to their inverses. Let $S=\lbrace g\in G\colon \phi(g)=g^{-1}\rbrace$.
Notice that if $g$, $h$ and $gh$ all lie in $S$ then on the one hand, $\phi(gh)=(gh)^{-1}=h^{-1}g^{-1}$. On the other hand, $\phi(gh)=\phi(g)\phi(h)=g^{-1}h^{-1}$, so that $g^{-1}$ and $h^{-1}$ commute. It follows that $g$ and $h$ commute.
Fix $g\in S$ and consider $A=\lbrace h\colon h\in S\text{ and } gh\in S\rbrace$. There are less than $|G|/4$ $h$'s for which the first condition fails and less than $|G|/4$ $h$'s for which the second condition fails, so that $|A|>|G|/2$. By the above, it follows that the centralizer of $g$ (i.e. the set of $h$'s that commute with $g$) is a superset of $A$. Since the centralizer is a subgroup, by Lagrange's theorem the centralizer of $g$ must be all of $G$. That is $g$ lies in the center of the group. Now we have that more than $1/2$ of the group lies in the center (which is again a subgroup), so that $G$ is Abelian.
Now since $S$ is more than half of the group, the subgroup generated by $S$ must be all of $G$, so that every element of $G$ is a product of elements of $S$. Now $g\in G$, write $g=s_1\ldots s_n$. Then $\phi(g)=\phi(s_1)\ldots\phi(s_n)=s_1^{-1}\ldots s_n^{-1}=s_n^{-1}\ldots s_1^{-1}=g^{-1}$, so that $\phi(g)=g^{-1}$ for all $g\in G$.
