I am learning about group actions and experimenting with different actions. One common action I see is acting by conjugation on the power set.
From this action we get the stabilizer of a set $A \subset G$ is its normalizer: $\{g \in G \mid gAg^{-1}=A\}=N_G(A)$.
The set of fixed points is $\{A \in \mathcal P(G) \mid gAg^{-1}=A \text{ for all } g \in G\}$. Can we say something more about this set? I see that if we instead act on the set of subgroups, then the set of fixed points is the set of normal subgroups.
The kernel is $\{g \in G \mid gAg^{-1}=A \text{ for all } A \in \mathcal P(G)\}$. So, $g \in N_G(A)$ for all $A \in \mathcal(G)$. Can we say something more about this set? Is the kernel just the union of all the normalizers? The kernel is normal and so it is the union of conjugacy classes. Which conjugacy classes?
For the set of fixed points, I don't think there's much you can generally say, but such points are sometimes called normal subsets of $G$. It's not hard to show that these are unions of conjugacy classes.
For the kernel, I'll give a hint. If $g$ is in the kernel, then for all $x\in G$ you have $\{x\}=\{x\}^g=\{x^g\}$. What does this tell you about $g$?