**Attempt:**
I don't know how to show that \pi (g) = g^{-1}, using the given hint. Any help to get me started is really appreciated. But I have tried to show that G is abelian and that it has an odd order:

If $\displaystyle \pi $ is an automorphism then for all $\displaystyle g,h \in G$

Hence G is Abelian. (Right?)

Now to show that the group is of an odd order. We can split the whole set G\{1} into disjoint pairs

.

Then G\{1} has even number of elements and thus $\displaystyle G$ (including 1) has an odd number of elements.

Is this right?

P.S. Latex wasn't working properly today.