An element of a group $G$ is not conjugate to its inverse if $\lvert G\rvert$ is odd Prove that if $G$ is a finite group of odd order, then no $x\in$$G$ , other than $x=1$, is conjugate to its inverse.


This question is from Advanced Modern Algebra (exer 2.79) by Joseph J. Rotman.

The hint states that if $x$ and $x^{-1}$ are conjugate, how many elements are in $x^{G}$?

What I know so far:


$\left\lvert x^{G}\right\rvert$ is odd (greater than 1, otherwise it's in the center) and is a divisor of |$G$|
|Z($G$)| has a common factor (other than 1) with size of the orbit $\left\lvert x^{G}\right\rvert$ so that the center is not just the identity. This is from the class equation.
The centralizer of $x$ has odd size and $\lvert C_{G}(x) \rvert \cdot \lvert x^{G}\rvert=\lvert G\rvert$


I don't see the implication of $x$ and its inverse being conjugate has other than their orbits having the same size and laying in the same conjugacy class.

Is the info I have useful? Any help would be appreciated.
Suppose $x \neq 1$ is conjugate to $x^{-1}$.
Let $C(x)$ be the conjugate class containing $x$.
If $x = x^{-1}$, then $x^2 = 1$.
Hence $|G|$ is divisible by $2$.
This is a contradiction.
Hence $x \neq x^{-1}$.
Since $|C(x)|$ is odd, $C(x)$ contains $y$ which is neither $x$ nor $x^{-1}$.
Since $x$ is conjugate to $y$, $x^{-1}$ is conjugate to $y^{-1}$.
Hence $y^{-1} \in C(x)$.
Since $y \neq y^{-1}$, we must conclude that $|C(x)|$ is even.
This is a contradiction.
$|G|$ is odd implies that no element (other that 1) has even order. Suppose $x$ conjugate to its inverse, then  $x=g(gx)^{-1}$ implies $gx=g^2(gx)^{-1}$, so we obtain that $g^2=1$, a contradiction is found.