Odd-order and ambivalent implies trivial

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

Facts used

 * 1) uses::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.