Odd-order and ambivalent implies trivial

Statement

Suppose $G$ is an Odd-order group (?) (i.e., a finite group of odd order) that is also an Ambivalent group (?): every element is conjugate to its inverse. Then, $G$ is the trivial group.

Facts used

1. Order of element divides order of group

Proof

Suppose $G$ has odd order, is ambivalent, and is nontrivial. Then, there exists a non-identity element $a$ in $G$. By fact (1), $a$ has odd order, so $a \ne a^{-1}$.

Since $G$ is ambivalent, there exists $b \in G$ such that $bab^{-1} = a^{-1}$. Then, $b^2ab^{-2} = a$, so $a$ and $b^2$ commute. Again by fact (1), $b$ has odd order, so $\langle b^2 \rangle = \langle b \rangle$. Since $a$ commutes with $b^2$, it must commute with all elements in $\langle b^2 \rangle$, and hence with $b$. Thus, $bab^{-1} = a$. This forces $a = a^{-1}$, a contradiction.