TITLE: Group element not taken to its inverse by any automorphism
QUESTION [13 upvotes]: What is an example of a group $G$ with an element $g$ such that no automorphism of $G$ takes $g$ to its inverse?

REPLY [7 votes]: The group $G=\mathbf{Z}/7\mathbf{Z}\rtimes\mathbf{Z}/3\mathbf{Z}$, where the generator $t$ of $\mathbf{Z}/3\mathbf{Z}$ acts by multiplication by 2: $t$ is not conjugate to $t^{-1}$ by an automorphism. Indeed the full automorphism group is $H=\mathbf{Z}/7\mathbf{Z}\rtimes\mathbf{Z}/7\mathbf{Z}^*$, which contains $G$ as subgroup of index 2. So the elements of $G$ conjugates to their inverse by an automorphism are only the elements of $\mathbf{Z}/7\mathbf{Z}$.
To check the assertion on the automorphism group, observe that if we have an automorphism of $G$, then after composing by an inner automorphism, it maps the 3-Sylow $\mathbf{Z}/3\mathbf{Z}$ into itself, and after composing by an element of $\mathbf{Z}/7\mathbf{Z}^*\subset H$, it acts as the identity on $\mathbf{Z}/7\mathbf{Z}$. So it is either the identity, or acts as identity on $\mathbf{Z}/7\mathbf{Z}=\langle u\rangle$ and maps $t$ to $t^{-1}$, but the latter does not define an automorphism, because it is not compatible with the conditions $tut^{-1}=u^2\neq t^{-1}ut$.