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?
 A: Define $\;G_T:=\{g\in G\;;\;Tg=g^{-1}\}\;$. Try now to prove the following:
$$\begin{align*}\bullet&\color{green}{x,y,xy\in G_T\implies xy=yx}\;\;\color{red}{Hint}:\text{calculate in two different ways}\;\;T(xy)\\{}\\
\bullet&\color{green}{\forall\,x\in G_T\;,\;G_T\cap x^{-1}G_T\subset C_G(X)}\;\;\color{red}{Hint:}\;y\in G_T\cap x^{-1}G_T\implies y=x^{-1}s\;,\;s\in G_T\implies\\{}\\&xy=s\;,\;\;\text{and now apply the first point above}\\{}\\
\bullet&\color{green}{\forall\,x\in G_T\;,\;\left|G_T\cap x^{-1}G_T\right|>\frac{|G|}2}\;\color{red}{Hint:}\left|G_T\cap x^{-1}G_T\right|=|G_T|+|x^{-1}G_T|-|G_T\cup x^{-1}G_T|\\{}\\&\text{and check that}\;\;|x^{-1}G_T|=|G_T|\\{}\\
\bullet&\color{green}{\forall\,x\in G_T\;,\;\;C_G(x)=G}\;\;\color{red}{Hint:}\;C_G(x)\le G\;\;\text{and the above}\\{}\\\bullet&\color{green}{G_T=Z(G)}\;\;\color{red}{Hint:}\;\text{immediate from the above}\\{}\\\end{align*}$$
Now you just need to deduce that $\;G\;$ is abelian (same argument as fourth point above and Lagrange's theorem), and thus $\;\langle\,G_T\,\rangle=G\;$ , which means that the endomorphism $\;T\;$ is inversion in a generating set of the group so...
Counterexample: take the quaternion group
$$Q_8=\{1,-1,i,j,k,-i,-j,-k\}$$
with the well known relations, and let $\;T\;$ and endomorphism s.t.
$$Ti=-i\;,\;\;Tj=-j\;,\;\;Tk=k$$
Observe that $\;T\;$ has been defined on a generating set of $\;Q_8\;$ so expand according to this, and check that exactly six elements are mapped to their inverses, yet the group's obviously not abelian and the map isn't inversion.
Something very similar can be done with the other non-abelian group of order $\8\;$, the dihedral group...
A: 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.

