Are Aut-Equivalent classes the same as Conjugacy Classes?

Consider a finite group $G$, elements $u,v \in G$ are said to be $\text{Aut}$ equivalent if $\exists \phi \in Aut(G) | \phi(u) = v$. You can then consider Aut-equivalence to be an equivalence relation leading to Aut-equivalent classes.

A restriction of this idea is to consider the "conjugacy" classes of a group. Which are defined by the conjugacy relations, said to be true for a pair $u,v \in G$ if $\exists g \in G \ s.t. \ gug^{-1} = v$.

Claim: The Aut-equivalent classes of a group are the conjugacy classes of a group

My work

1. Let $\phi$ be an arbitrary automorphism. Then if $u,v$ are conjugate meaning $\exists g \ gug^{-1} = v$ it follows that $\phi(g)\phi(u)\phi(g)^{-1} = \phi(v)$ that is $\phi(u), \phi(v)$ are conjugate.

2. Suppose there doesn't exist $t$ such that $tut^{-1} = v$ but there if there exists $t$ such that $t \phi(u) t^{-1} = \phi(v)$ Then it must be the case that $\phi^{-1}(t) u \phi^{-1}(t)^{-1} = v$. That is if 2 elements are conjugate post automorphism, then they must have been conjugate prior.

This tells me that under any automorphism, conjugacy is respected. Which is a start... but doesn't really tell me anything more useful.

Another aspect is that for every Aut-equivalent class $A$ there exists one more entire conjugacy classes $C_i$ tha $C_i \subseteq A$, in other words Aut-equivalent classes can only be larger than a union of some finite number conjugacy classes.

Furthermore conjugation is an automorphism, so the "larger" statement can be made stronger into

Every Aut-equivalent classes is a finite union of some number of conjugacy classes.

But going from finite to 1... feels like a jump

• Have you tried looking at what happens in examples? For instance the cyclic group of order $3$? – Sebastian Schoennenbeck Mar 2 '17 at 7:51
• oh... {0}, {1}, {2} form their own conjugacy classes but here there's clearly an automorphism that combines them. I'm silly – frogeyedpeas Mar 2 '17 at 7:53
• On the positive side: You learned something about groups, and everything else you did is perfectly fine, it just doesn't lead to the desired conclusion. – Sebastian Schoennenbeck Mar 2 '17 at 7:57
• And, hopefully, you've also learned that you have to consider examples. – Mariano Suárez-Álvarez Mar 2 '17 at 8:09

For each $g \in G$ the map $G \to G$ of the form $x \mapsto gxg^{-1}$ is called an inner automorphism of $G$. It is a simple exercise to check this map is indeed an automorphism. Every inner automorphism is uniquely determined by the element $g$.
Observe that all inner automorphisms on an Abelian group are trivial, and indeed the conjugacy classes are trivial. So our best bet is to find an Abelian group $A$ with a nontrivial automorphism $h \colon A \to A$. Then any pair $a$ and $h(a)$ will be Aut-equivalent but not conjugate unless $a = h(a)$.