Isotropy group of a totally geodesic submanifold in a Riemannian homogeneous space acts transitively on the submanifold I was wondering if the following holds:

Let $G/H$ be a (complete) Riemannian homogeneous manifold (i.e., a homogeneous manifold with a prescribed Riemannian metric on it, and $G$ acts transitively on $G/H$ through isometries). Suppose $\xi_0$ is a totally geodesic submanifold of $G/H$, and let $H_0 = \left\lbrace g \in G | g \cdot \xi_0 = \xi_0 \right\rbrace$ be its isotropy group. Can we say that $H_0$ acts transitively on $\xi_0$?

It seems to be true in simple cases of $\mathbb{R}^n$ and $\mathbb{S}^n$. In the first case, $G$ is the group of rigid motions characterized by the matrices of the type $\left[ \begin{matrix} A & b \\ 0& 1 \end{matrix} \right]$, where $A \in O(n)$ and $b \in \mathbb{R}^n$, with the action given by $\left[ \begin{matrix} A & b \\ 0& 1 \end{matrix} \right] \cdot x = Ax + b$, for $x \in \mathbb{R}^n$. Here, if I consider $\mathbb{R}^{n - 1}$ as a totally geodesic submanifold of $\mathbb{R}^n$, we get its isotropy group to be
$$H_0 = \left\lbrace \left[ \begin{matrix} A & 0 & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] : A \in SO(n-1) \text{ and } b \in \mathbb{R}^{n - 1} \right\rbrace \cup \left\lbrace \left[ \begin{matrix} A & 0 & b \\ 0 & -1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] : A \in O^{-}(n-1) \text{ and } b \in \mathbb{R}^{n - 1} \right\rbrace,$$
where $O^{-}(n-1)$ is the set of all orthogonal matrices with determinant negative. From this characterization, it is easy to see that $H_0$ does act transitively on $\xi_0$ (through translations alone!). Similar computations can be done for the case of $\mathbb{S}^n$, where the totally geodesic submanifolds are lower dimensional spheres.
Now, I want to generalize this to arbitrary Riemannian homogeneous manifolds (that need not be embedded in any Euclidean space, like $\mathbb{S}^n$). I have an intuition that this should be true. That is because, just like the case of $\mathbb{R}^n$, there are group elements from $G$ that take one point to another, and then there are elements from $H_0$ and do not take us outside $\xi_0$. Therefore, it seems that by choosing proper element, we will remain inside $\xi_0$ and yet be able to go from one point to another using elements of $H_0$.
Any help in formalizing these ideas (or disproving them) will be highly appreciated!
 A: In general, no, there are totally geodesic subgroups which are not acted on transitively by the group.
To see this, consider the homogeneous spaces $SU(n+1)/SU(n)$, which happens to be diffeomorphic to the sphere $S^{2n+1}$.  Here, for definiteness, I'm thinking of the $SU(n)$ as being embedded in $SU(n+1)$ as matrices of the block diagonal form $\operatorname{diag}(1,A)$ with $A\in SU(n)$.
Let $\xi_0$ denote the equatorial $S^{2n}\subseteq S^{2n+1}\subseteq \mathbb{C}^{n+1}$ given by $S^{2n} = \{(z_1,...,z_{n+1})\in S^{2n+1}: Im(z_1) = 0\}$.  Then $\xi_0$ is totally geodesic, being the fixed point set of the isometry $\phi:S^{2n+1}\rightarrow S^{2n+1}$ given by $\phi(z_1,...,z_{n+1}) = (\overline{z}_1, z_2,..., z_{n+1}).$
Then, for any $n\geq 1$, the isotropy group $H_0$ does not act transitively on $\xi_0$.
Here is one way to see it.  From the explicit description of $SU(n)$ above, it follows that $SU(n)\subseteq H_0$.  The subgroup $SU(n)\subseteq SU(n+1)$ is almost maximal (among connected groups): it is contained in $U(n)$, but no larger group.  Thus, the identity component of $H_0$ (which acts transitively on $\xi_0$ iff the whole $H_0$ action on $\xi_0$ is transitive) is either $SU(n)$ or $U(n)$.  But from the classification of transitive actions on a sphere, there is no transitive action of $SU(n)$ or $U(n)$ on $S^{2n}$.
In the example above, a bi-invariant metric on $SU(n+1)$ does not induce the round metric on $S^{2n+1}$, so one may still ask for a normal homogeneous space where this kind of thing happens.  Here is an example.
Consider the homogeneous space $G_2/SU(3)$, which is diffeomorphic to $S^6$.  Here, a bi-invariant metric on $G_2$ does induce the round metric on $S^6$.  Now, let $\xi_0$ denote any equatorial $S^4$.  From the classification of transitive actions on spheres, the only group acting transitively on $S^4$ is (up to covers) $SO(5)$.  But there is no $SO(5)\subseteq G_2$.
