A corollary of the slice theorem I'm going to prove the corollary $5.6.26$ in Riemannian Geometry of Peter Petersen:
Corollary 5.6.26. If $\mathrm{H} \subset$ Iso $(M, g)$ is a closed subgroup with the property that all its isotropy groups are conjugate to each other, then the quotient space is a Riemannian manifold and the quotient map a Riemannian submersion.
To prove this, I'm trying to use the free slice theorem, i.e.
Theorem 5.6.21 (The Free Slice Theorem). If $\mathrm{H} \subset$ Iso $(M, g)$ is closed and acts freely, then the quotient $\mathrm{H} \backslash M$ can be given a smooth manifold structure and Riemannian metric so that $M \rightarrow \mathrm{H} \backslash M$ is a Riemannian submersion.
So we just need to prove that for any $p\in M$, $H_p=\{e\}$. I use corollary $5.6.25$:
Corollary 5.6.25. For small $v \in T_p^{\perp} \mathrm{H} p$ the isotropy at $\exp _p(v)$ is given by
$$
\mathrm{H}_{\exp _p(v)}=\left\{h \in \mathrm{H}_p|D h|_p v=v\right\}
$$
Thus by the assumption, we have for small $v \in T_p^{\perp} \mathrm{H} p$, $\mathrm{H}_{\exp _p(v)}=H_p$. Therefore for any $h\in H_p$, $Dh_p|_{T_p^{\perp} \mathrm{H} p}=id$.
But to show $H_p=\{e\}$, I have to show $Dh_p|_{T_p \mathrm{H} p}=id$. I don't know how to solve it.
 A: If you look over Petersen's proof in the free case, you'll see that one of the key steps in the proof is establishing that an $H$-orbit $H\cdot p$ (which is diffeomorphic to $H$) has an $H$-invariant neighborhood which is equivariantly-diffeomorphic to $H\times B$ where $B\subseteq T_p^\bot (H\cdot p)$ is a small ball centered at the origin.
For the non-free case, but where all isotropy groups are conjugate, the corresponding result is
Theorem:  Every $H$-orbit $H\cdot p$ has an $H$-invariant neighborhood equivariantly diffeomorphic to $(H/H_p) \times B$, with $B$ as above.
Once one has this result, Petersen's proof in the free case carries over almost verbatim to this case.
So, let's prove the Theorem.
Using Theorem 5.6.24, we see the an orbit $H\cdot p$ has an $H$-invariant neighborhood which is equivariantly diffeomorphic to $(H\times_{H_p} B$ with $B$ as above. (This notation refers to the quotient $(H\times B)/H_p$ where $H_p$ acts on $H$ via multiplication and on $B$ via the isotropy action: for $h\in H_p$ and $v\in B\subseteq T_p M$, $h\ast v = (Dh|_p) v$).
In general, there is no reason the $H_p$ action on $B$ should be trivial, but...
Proposition If all isotropy groups are conjugate, then the $H_p$ action on $B$ is trivial.
Believing this proposition momentarily, it now follows that $(H\times B)/H_p = (H/H_p)\times B$, so we're done.
So, why is the Proposition true?  Well, select $v\in B$.  Corollary 5.6.25 tells us that $H_{\exp{v}} = \{h\in H_p: Dh|_p v = v\}$, so $H_{\exp(v)}$ is obviously a subgroup of $H_p$.  On the other hand, by assumption, $H_{\exp(v)}$ and $H_p$ are conjugate.  Conjugation is a diffeomorphism, so $H_p$ and $H_{\exp(v)}$ are the same dimension, and have the same number of components.  Moreover, isotropy groups of isometric actions are always compact, so the number of components must be finite.
We claim that this implies that $H_{\exp(v)} = H_p$.  To see this, note that both are compact submanifolds of the same dimension.  So, the inclusion $H_{\exp(v)} \subseteq H_p$ automatically implies that $H_{\exp(v)}$ is a union of components of $H_p$.  But since the number of components match, we must have $H_p = H_{\exp(v)}$.
But now, we see that $$H_p = H_{\exp(v)} = \{h\in H_p: Dh|_p v = v\},$$ so we conclude that $Dh|_p v = v$ for all $h \in H_p$.  Said another way, $H_p$ acts trivially on $B$, as claimed.  This completes the proof of the Proposition, and hence, of the Theorem.
