TITLE: Groups with "unique" elements of high order
QUESTION [11 upvotes]: A group $G$ can have an element $g$ for which every automorphism of the group fixes $g$. Obviously, the identity is one such element, and one can easily find order-2 examples: the unique order-2 element in $C_{2n}$, or $-1$ in the quaternion group.
My question is whether one can find a group $G$ with an element $g$ of order at least 3 which is fixed by every automorphism of $G$.
One might suspect that there is always an automorphism taking $g$ to $g^{-1}$, presenting an obvious obstacle to the high-order case, but this is false. For example, take $G$ to be the unique nonabelian group of order $21$, realized as a semidirect product of $C_7$ and $C_3$. Then any of the $14$ elements of order $3$ cannot be sent to their inverses. However, none of these elements are fixed by every automorphism; the $14$ elements fall into two conjugacy classes of size $7$.
Restricting our attention to those automorphisms given by conjugation, we see that $g\in Z(G)$. However, I haven't found a way to strengthen this restriction into a proof of impossibility.
Edit: This post originally contained the sentence

Intuitively, one can think of such an element as "unique" in the sense that it has group-theoretic properties not shared by any other element.

but I have moved it to the bottom to avoid confusion. By this, I just mean that automorphisms exchange elements that in some sense "serve the same role" as each other in the group; an element which is fixed by every automorphism can be thought of as having no such counterparts. I do not mean to search for a group in which only one element has the property of being fixed by every automorphism.

REPLY [9 votes]: I think the question is asking for a finite group having an element of order at least 3 that is fixed by every automorphism. The smallest such group has order 63 and is the unique non-trivial semidirect product $C_7\rtimes C_9$.
Here's a sketch of why it works. Let $G=C_7\rtimes C_9$. First, it's not too hard to see that the center of $G$ has order $3$. We claim that elements of the center are fixed by every automorphism.  Now, since $Z(G)$ is characteristic in $G$, $\mathrm{Aut}(G)$ has an induced action on $G/Z(G)$, which is the non-abelian group of order $21$. As the OP pointed out, in this group, an element of order $3$ and its inverse are in different orbits of the automorphism group. It then follows that, if we  pull back the action to the full $G$, the inverse pair of elements of order $3$ in $Z(G)$ are also in different orbits.