Let $(M,\mu)$ be a Riemannian manifold with an isometric action of a compact Lie group $G$. If the action is not free, the orbit space $M/G$ is a stratified space \begin{equation} M/G = \bigcup_{H<G} M_{(H)}/G \end{equation} where \begin{equation} M_{(H)}:= \{ x\in M \,| \, G_x = g H g^{-1} \text{ for some } g\in G\} \end{equation} This is a submanifold of $M$. Consider another submanifold contained in $M_{(H)}$ defined by, \begin{equation} M_H := \{ x \in M \,|\, G_x =H \} \end{equation} Both $M_{(H)}$ and $M_H$ inherit a Riemannian metric from the ambient manifold $M$.
The slice theorem tells us that the orbit space $M_{(H)}/G$ is a manifold. Denoting $\pi_{(H)}: M_{(H)} \rightarrow M_{(H)}/G$. Then the isomorphism, \begin{equation} d\pi_{(H)} : (T_x{(G\cdot x)})^{\perp M_{(H)}} \xrightarrow{\cong} T_{\pi_{(H)}(x)}(M_{(H)}/G) \end{equation} can be used to define a Riemannian metric on the orbit space, denoted by $\mu_{(H)}$, turning the projection into a Riemannian submersion.
On the other hand, on the submanifold $M_H$, we have the action of the group $N_G(H)$ which is the normaliser of $H$ in $G$ (maximal subgroup of $G$ in which $H$ is a normal subgroup). Then the quotient group $L:=N_G(H)/H$ acts freely on $M_H$ and $M_H/L$ is a manifold as well. We can define a metric $\mu_H$ on $M_H/L$ to make the natural projection map from $M_H$ into a Riemannian submersion.
Now as manifolds, we have a diffeomorphism \begin{equation} M_H/L \cong M_{(H)}/G \end{equation}
Question:
- Are the Riemannian manifolds $(M_H/L, \mu_H)$ and $(M_{(H)}/G, \mu_{(H)}) $ isometric?
- If not, then are there some group actions where the above are isometric? One trivial case is when $H<G$ is a normal subgroup in which case $M_{(H)}=M_H$