A dodecahedron out of five tetrahedra AKA Partitioning an orbit in X/K into orbits under cosets of H in K Consider the orbit space $X/K$ with $X$ a symmetric space and $K$ a group. Let $x$ represent an orbit $Kx$ in $X/K$. Now let's introduce a subgroup $H \subset K$, split up $K$ into cosets $aH$ (with $a \in K/H$), and look at the parts of the orbit corresponding to each coset, i.e., $aHx$. It seems that the coset representative $a$ either rotates $Hx$ to a distinct subset of $X$ or leaves it be. In other words, are there any weird situations where only some points of the coset orbits $aHx$ and $bHx$ are the same but others are different? 
I know of one famous example, the compound of five tetrahedra, in which five tetrahedra (corresponding to the five coset orbits $aTx$) together make up a dodecahedron $Ix$ (an orbit in $S^2/I$). So I was wondering what happens in the general/generic cases for $S^2$ and $H \subset K \subset SO(3)$.
 A: Let $Kx$ be some orbit in the orbit space $S^2/K$. Consider the set of points $Hx$, i.e., the orbit of $x$ under $H$. The stabilizer of the set $Hx$, $\text{stab}(Hx)=H$ generically (we'll do general case later). Applying the orbit-stabilizer theorem, we can then determine the number of elements in the orbit of $Hx$ under $K$: 
$$
|\text{orb}(Hx)|=|K|/|\text{stab}(Hx)|=|K|/|H|.
$$
Thus, each element of the orbit of $Hx$ under $K$ is in one-to-one correspondence to a coset $aH$ of $H$ in $K$.
Let's come back to the non-generic cases. Generally, $\text{stab}(Hx)$ must be some subgroup $P$ such that $H\subseteq P \subseteq K$. So if, e.g., $H$ is the largest subgroup of $K$, then $P$ is either $H$ (generic case above) or $K$ (case at a few special points).
For the tetrahedral compound case, $H=T$ and $K=I$, and we'll have $|I|/|T|=5$ elements in the orbit of $Tx$ under $I$ for almost all $x$. An exception is when $x=x^\star$ is a vertex of the icosahedron that is symmetric under $I$. That icosahedron has 12 vertices, which can be obtained by applying $T$ to $x^\star$. Then, $Tx^\star=Ix^\star$ and we have $|I|/|I|=1$ element only. Since there is no subgroup inbetween $T$ and $I$, the above two cases are the only ones that can happen.
