The question is as such:
It's given that |G| is odd, and there is a fixed element a in G. I need to prove that for every element g in G, there is a unique equation g = xax for some x in G (where x is unique for each corresponding g).
The only possible helpful step I have is knowing that every element (hence g, a, and x) is of odd order. I haven't found a way to apply this, so I'm not entirely sure if it's useful. This seems like it'd be easy if the group was abelian, and hence commutative, but this isn't stated (potentially it can be proved?).
Equivalently, you're trying to solve $ag = (ax)^2$. For any element $y$ such that $y^2 = ag$, setting $x = a^{-1}y$ solves the equation (and vice versa), and so really, we are trying to show that every element of $G$ has a unique square root.
To take a square root of a $g$, we can define $\sqrt{g} := g^{\frac{|G|+1}{2}}$: then $(\sqrt{g})^2 = g^{|G|+1} = g$ since we know $g^{|G|}$ is the identity. This is unique: if we suppose that $g^2 = h^2$, then by raising both sides to the $\frac{|G|+1}{2}$ power, we get $g^{|G|+1} = h^{|G|+1}$ and conclude $g=h$.
Therefore every element has a unique square root, and $g = xax$ is solved by setting $x = a^{-1}\sqrt{ag}$.