What happens when we partition G into equivalence classes induced by its automorphism group?

Question

Imagine we partition G into its orbits under the action of Aut(G) ($O_a=\{\phi(a) | \phi \in Aut(G)\}$) and consider each orbit as an element with the operation being the same as G (so $O_a •O_b=O_{ab}$). If I’m not wrong, this should be a group. Is this group interesting in any way? I can’t really put into words how/why I came up with this, I just thought that if we draw the lines of symmetry in a shape and take one of the subregions it gives us information about the shape as a whole, and I think what I did translated that idea from algebra to groups. I’m very rusty on group theory so I apologize if any part of this was phrased incorrectly or if I overlooked something obvious. Thanks for any help!

Answer 1

Your proposed operation $$O_a,O_b\mapsto O_{ab}$$ is not in fact always well-defined. Consider for example the group of rationals, $\mathbb{Q}$, with respect to addition. This group has only two orbits, namely $\mathbb{Q}\setminus \{0\}$ and $\{0\}$. This means that for example $$O_1=O_{-1}\mbox{ but }O_{1+(-1)}\not=O_{1+1}.$$

More generally, if $G$ is an abelian group which has an element $a$ which is neither the identity nor its own inverse, we will have $$O_a=O_{-a}\mbox{ but }O_{a+a}\not=O_{a+(-a)}.$$