Given an open subset of a Riemannian manifold, is the group of isometries that extend to a global isometry closed? Let $M$ be a (smooth) complete Riemannian manifold with isometry group $G$, which is a Lie group equipped with the compact-open topology (by Myers-Steenrod).
Let $U \subseteq M$ be a small open geodesic ball with center $p$. Consider the group of isometries $H = \operatorname{Isom}(U)$ and the stabilizer $H_p$. We have continuous restriction morphisms $G \to H$ and $\rho : G_p \to H_p$. Because $M$ is complete, they are injective.

Is the image $\rho(G_p)$ closed? That is, is the subgroup of isometries of $U$ that fix $p$ and that extend to isometries of $M$, closed in $H$?

Note that $G_p$ is always closed in $G$, by definition of compact-open topology. Also, $G$ is closed in the diffeomorphism group. So the statements:


*

*$\rho(G_p)$ is closed in $H_p$

*$\rho(G_p)$ is closed in $H$

*$\rho(G_p)$ is closed in $\operatorname{Diff}(U)$


are equivalent.

Motivation. I'm trying to prove that $G_p$ is compact. (Or to find conditions under which this holds.) For $M$ complete, I reduced the compactness of $G_p$ to the above question. (They are equivalent.)
I don't mind assuming that $M$ is homogeneous, isotropic or symmetric.
 A: This is true. I found the answer in the slides: A. Isaev, Proper Group Actions in Complex Geometry: http://web.maths.unsw.edu.au/~aji23/Isaev071107.pdf
The stabilizers are compact for any Riemannian manifold. This follows from:

Theorem 1. (van Dantzig, van der Waerden 1928) If $(M, d)$ is a connected locally compact metric space, then the group $\operatorname{Isom}(M)$ of all isometries of $M$ with respect to $d$ (considered with the compact-open topology) is a topological group acting properly on $M$ by homeomorphisms.

The important fact is that the action is proper. We conclude using:

Theorem 2. (Myers, Steenrod) The isometry group of a Riemannian manifold acts properly.

There are different nonequivalent notions of proper group actions; in this case (everything is locally compact Hausdorff) they are all equivalent and properness can be taken to mean that the map:
$$G \times M \to M \times M$$
$$(g, p) \mapsto (gp, p)$$
is proper. In particular, stabilizers (=isotropy subgroups) are compact.

For completeness (ha ha), here is how this is equivalent to the question. If $G_p$ is compact, then certainly its image $\rho(G_p)$ is closed (because compact).
Conversely, suppose it is closed. The exponential map $\exp_p$ gives a topological group isomorphism
$$\operatorname{Diff}(U) \to \operatorname{Diff}(\exp^{-1}(U))$$
and hence identifies the closed subgroup $H_p \subseteq \operatorname{Diff}(U)$ with a closed subgroup of the compact group  of orthogonal transformations $O(T_pM)$ restricted to $\exp^{-1}(U)$. Note here that restriction to $\exp^{-1}(U)$ gives a homeomorphism on its image $O(T_pM) \to \operatorname{Diff}(\exp^{-1}(U))$. This is because the compact-open topology on $O(T_pM)$ coincides with the norm topology (both are the topology of locally uniform convergence) and this coincides with the induced compact-open topology on its image in $operatorname{Diff}(\exp^{-1}(U))$, because we can recover the norm of a linear operator by only looking at its action on a neighborhood of $0$.
Finally, because all those groups are closed, they are Lie groups so the continuous group homomorphism $G_p \to O(T_pM)$ is a homeomorphism on its closed image. (We assume $M$ is complete for this map to be injective.)
