If i am able to prove that "if $\displaystyle g$ is in the orbit of $\displaystyle x$ then so is $\displaystyle g^{-1}$" then i will be proving the opposite of what the question asks me to prove. Because then i will say "since $\displaystyle x$ is in the orbit of $\displaystyle x$ then so is $\displaystyle x^{-1}$".... which will mean that $\displaystyle x$ and $\displaystyle x^{-1}$ are in the same orbit and this is the opposite of what i have to prove.

It can however be said with certainty that $\displaystyle g \neq g^{-1}$ for each $\displaystyle g \neq 1$ since the order of the group is odd.

Please consider this. Thanks.