Let $G$ be a real Lie group.
What conditions must $G$ satisfy so that the following is true:



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> For any finite group $\Gamma$ there exist finitely many conjugacy classes of subgroups of $G$ that are isomorphic to $\Gamma$.
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I believe that for $G=GL(n,\mathbb{R})$ this is true because:
subgroups of $GL(n,\mathbb{R})$ are conjugate if and only if the restriction of the standard representation of $GL(n,\mathbb{R})$ to the subgroups are isomorphic representations.
Finite groups have finitely many irreducible representations, and that proves the claim.


I also believe that for $G=SO(n)$ this is true because of this answer: <https://mathoverflow.net/a/17074/164084>.


It would be enough for my application to know that every real, compact, connected Lie group that has a faithful representation has the property stated above ("For any finite group $\Gamma$...").


This is a crosspost from Stackexchange Mathematics, see here: <https://math.stackexchange.com/questions/4219091/which-lie-groups-have-finitely-many-conjugacy-classes-of-subgroups-of-fixed-isom>.



Which Lie groups have finitely many conjugacy classes of subgroups of fixed isomorphism type?
A natural condition is that $G$ has finitely many connected components.
One can easily reduce this case to the connected group case, and then to the compact group case, as all [maximal compact subgroups](https://en.wikipedia.org/wiki/Maximal_compact_subgroup) in a connected Lie group are conjugated.
Then the representation variety $\text{Hom}(\Gamma,G)$ is compact and [local rigidity](https://en.wikipedia.org/wiki/Local_rigidity), that is vanishing of $H^1(\Gamma,\mathfrak{g})$, guarantees the finiteness of the number of $G$-orbits. Here $\mathfrak{g}$ denotes the Lie algebra of $G$. The fact that $H^1$ vanishes could be deduced from the fact that every isometric action of $\Gamma$ on $\mathfrak{g}$ has a fixed point, by averaging an orbit.


