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Proof. Let $Y:={g \in G: Tg=g^{-1}}$. Then $|Y|>\frac{3}{4}|G|$ by hypothesis. Choose $a \in Y$ and set $aY=:Z$. It's easy to see that $|Z|=|Y|$. Now we want to compute $|Y cap Z|$. Since $|G setminus Y|=|G setminus Z| \le 1/4|G|$, simply using the De Morgan laws, we get $|Y cap Z| > \frac{|G|}{2}$. Now pick $s in Y cap Z$ and $y \in Y$: $s$ must be of the form $ay$ and since $s \in Y$, $y^{-1}a^{-1}=T(ay)=T(a)T(y)=a^{-1}y^{-1}$, i.e. $y \in C(a)$ (with $C(cdot)$ we've indicated the centralizer of $cdot$).
Now remember that $C(cdot)$ is a subgroup, and in particular we have just shown that $forall a in Y$, $|C(a)|>|G|/2$; by Lagrange's theorem, $C(a)=G$. This means that every element of $Y$ commutes with all the element of the group: in other words, $Y subseteq Z(G)$. But $3/4|G| <|Y| \le |Z(G)|$ which gives $Z(G)=G$ so $G$ is abelian.
In this case, we can easily prove that $Y$ is indeed a subgroup and, since it has got more than $3/4$ of the elements of $G$, it must be $G$ itself, i.e. $Tx=x^{-1}, forall x in G$.
Question. What if $Tx=x^{-1}$ for exactly $3/4$ of the elements of $G$?
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