<s>[INST] How is this automorphism an inversion? And how is this group abelian? Let $G$ be a finite group, and let $T$ be an automorphism of $G$ which sends more than three-quarters of the elements of $G$ onto their inverses. Then how to demonstrate that $T(x) = x^{-1}$ for all $x$ in $G$? And, how to show that $G$ is abelian? 
Can we find an example of a finite, non-abelian group with an automorphism which maps exactly three-quarters of the elements of the group onto their inverses?
 [/INST] I am reading "Topics in Algebra 2nd Edition" by I. N. Herstein.
These problems are Problem 12 and Problem 13 on p.71 in Herstein's book.
I solved Problem 13 as follows:

I used GAP and I found the following example.
Let $H:=\{(), (1\text{ }2), (3\text{ }4), (1\text{ }2)(3\text{ }4), (1\text{ }3)(2\text{ }4), (1\text{ }4)(2\text{ }3), (1\text{ }3\text{ }2\text{ }4), (1\text{ }4\text{ }2\text{ }3)\}$.
Then, $H$ is a non-abelian subgroup of $S_4$. (I believe GAP.)
The identity mapping $\operatorname{id}$ is obviously an element of $\operatorname{Aut}(H)$.
And,
$()=()^{-1}$.
$(1\text{ }2)=(1\text{ }2)^{-1}$.
$(3\text{ }4)=(3\text{ }4)^{-1}$.
$(1\text{ }2)(3\text{ }4)=[(1\text{ }2)(3\text{ }4)]^{-1}$.
$(1\text{ }3)(2\text{ }4)=[(1\text{ }3)(2\text{ }4)]^{-1}$.
$(1\text{ }4)(2\text{ }3)=[(1\text{ }4)(2\text{ }3)]^{-1}$.
$(1\text{ }3\text{ }2\text{ }4)\neq(1\text{ }4\text{ }2\text{ }3)=(1\text{ }3\text{ }2\text{ }4)^{-1}$.
$(1\text{ }4\text{ }2\text{ }3)\neq(1\text{ }3\text{ }2\text{ }4)=(1\text{ }4\text{ }2\text{ }3)^{-1}$.
So, the identity mapping maps exactly three-quarters of the elements of $H$ onto their inverses.

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