Prove that if $x$ is conjugate to $x^{-1}$ and $y$, that $y$ is conjugate to $y^{-1}$ The full question: Prove the following:
a) If $x$ is conjugate to $x^{-1}$ and $y$, then $y$ is conjugate to $y^{-1}$
b) If $x$ is conjugate to $x^{-1}$ in a finite group, $G$, and $x \neq x^{-1}$, then the conjugacy class of $x$ has an even number of elements.
c) If $G$ has odd order, and $x \in G$ is not the identity element, the $x$ is not conjugate to $x^{-1}$.
So far I have:
a)
$x = g^{-1}x^{-1}g$ for some $g$
and $x = h^{-1}yh$ for some $h$  
Therefore $y = hxh^{-1}$ and so $y$ is conjugate to $x$.
Now, I presume the theory is that because $x$ is conjugate to its inverse, then $y$ is conjugate to its inverse - but how do I prove this?
b) using part a, each element also has it's inverse in the conjugacy class and therefore there is an even number of elements.
c) I'm not sure how to prove it, but it would use the idea that because it contains an odd order of elements,  it cannot include the pairs of element and inverse so therefore $x$ is not conjugate to $x^{-1}$

From $h^{-1}yh=x=g^{-1}x^{-1}g$ we obtain by inverting $h^{-1}y^{-1}h=g^{-1}xg=g^{-1}(h^{-1}yh)g$, which finally gives
$$
y^{-1}=hg^{-1}h^{-1}yhgh^{-1}=(hgh^{-1})^{-1}y(hgh^{-1}).
$$
Part c) has been shown already in this MSE quation.
