An automorphism of a finite group which sends more than three quarters of elements to their inverses Question is :
Let $G$ be a finite group and suppose the automorphism $T$ sends more than three quarters of elements of $G$ onto their inverses.
Prove that $T(x)=x^{-1}$ for all $x\in G$ and that $G$ is abelian.
What I could see is...
Once I prove $T(x)=x^{-1}$ for all $x\in G$ then I would have that $G$ is abelian.
Because, for any $a,b\in G$ we have 
$b^{-1}a^{-1}=(ab)^{-1}=T(ab)=T(a)T(b)=a^{-1}b^{-1}$
Thus, $ab=ba$ for all $a,b\in G$ Thus $G$ is abelian.
So, Only problem is to prove that $T(x)=x^{-1}$ for all $x\in G$
I am unable to use the fact that 

"$T$ sends more than three quarters of elements of $G$ onto their inverses"

May be I should take  : $A=\{x\in G : T(x)=x^{-1}\}$ and see if this is a subgroup (It is Not) or something like that.
Please help me to clear this.
I am very excited about this question So please do not spoil my excitement by posting full answer at once (This is a request cum order).
Please help me to do this by myself. 
Thank you so much :)

This is an old problem from Herstein, and IMHO, hard to solve without some hints : 
Let $A = \{x \in G : T(x) = x^{-1}\}$, and you want to prove that $A = G$ (and thus conclude that $G$ is abelian).
Take $a\in A$, and consider $K = A\cap a^{-1}A$. Given what you know about $|A|$, check that
$$
|K| \geq |G|/2
$$
For any $x\in K$, since $x \in A$ and $x\in a^{-1}A$, check that $xa = ax$. Conclude that $K \subset C(a)$.
This is true for each $a \in A$. 
Now what can you say about $Z(G)$?
