I have an isometry group $G$ which acts on units of a ring $R$ (not neessarily the full isometry group). I know the set $P_f$ of fixed points in $R$ of one member $f$ of $G$. Can I use the group rules of $G$ to find the corresponding fixed points $P_g$ in $R$ of another element $g$ of $G$ from the $P_f$ I have? If so, how?
Find $P_g=\{x\in R:gx=x\}$ from $P_f=\{x\in R:fx=x\}$
My Attempt
This is as much as I can work out: I've convinced myself that since $G$ is a group of homeomorphisms, then by topological conjugacy every member of $G$ should have the same number of fixed points as the others. Therefore the question has a solution. Maybe that's wrong?
At first I thought this would be easy, just take the group member that takes one fixed point to another and I would be done. But I quickly realised it doesn't work like that.
So with a bit of head scratching I concluded I need to find the homeomorphism that topologically conjugates ones isometry to the other. It's not immediate (to me at least) that this will be a member of $G$ although it should be a member of the full isometry group on $R$. Am I barking up the right tree?
My next thought was that any subgroup of the isometry group has gotta generate a whole conjugacy group or something that's closed under conjugacy and that it's going to become complicated, but there is probably a method of taking $f$ and $g$ and generating that supergroup of conjugacies I'm talking about.
Edit
I'm coming round to the view that isometries are going to fall into conjugacy classes and that for me to expect them to have corresponding fixed points, they will need to be in the same conjugacy class. Let's assume $f,g$ are in the same conjugacy class for the purposes of this question as I believe that to be the case in my specific example anyway.
If the question is based upon the premise that the isometries are topologically conjugate to each other and therefore they must share the same numbers of fixed points, then the key is to find the conjugating map between them. That is the homeomorphism $h$ such that $hfh^{-1}=g$.
Incidentally, if the group is commutative (as it turned out my specific case is) the only conjugating element of the group is identity, so any suitable conjugating function would lie outside of the commutator.