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.



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> 1) Can we construct all such groups?
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(it is asserted in comment to [the answer below](https://mathoverflow.net/a/84844/14094) that these are precisely those finite groups whose center has index 4)



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> 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?
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($a/b=1$ is achieved precisely for the inversion map on an abelian group; otherwise $a/b\le 3/4$ according to [the answer below](https://mathoverflow.net/a/84844/14094))



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> 3) Also, can these questions make sense in infinite groups?
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Finite groups with automorphism mapping $a/b$ of the elements of $G$ to their own inverses? Case $a/b=3/4$?
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 $g\in G$.


