Given a structure (of pretty much any type) $\mathfrak{A}$, there is a natural "sameness" relation on the elements of $\mathfrak{A}$: we set $a\sim b$ iff there is an automorphism, in whatever sense is appropriate, of $\mathfrak{A}$ sending $a$ to $b$. This is often called the orbit equivalence relation, and its classes are called orbits.
I'm interested in the following a priori coarsening of the orbit relation: set $a\approx b$ iff there are homomorphisms (again, in whatever sense is appropriate) $f,g$ with $f(a)=b$ and $g(b)=a$. In general it's not hard to see that $\approx$ can be much coarser than $\sim$, even when $\mathfrak{A}$ is finite. However, I don't know of any particularly natural examples of finite structures where $\sim$ and $\approx$ separate.
Question: Is there a finite group $G$ such that $\sim$ and $\approx$ do not coincide in $G$?
If $G$ is abelian, then the answer is negative by brute-force application of the classification of finite abelian groups. However, I don't see any way to extend this to non-abelian groups. I vaguely recall seeing that the answer remains negative for all groups, but that was long ago and I can't recall how the proof (supposedly) went.
Separately, I'm interested in any sources on $\approx$. In particular, does it too have a snappy name?
Best Answer
Let $G = C_2 \times S_3$, with $a\in C_2$ and $b \in S_3$ having order $2$. Then there is no automorphism of $G$ mapping $a$ to $b$ (since $a \in Z(G)$ and $b \not\in Z(G)$), but it is easy to define homomorphisms $f,g: G\to G$ with $f(a)=b$ and $g(b)=a$.