Quotient of HP^n by SO(3) = Aut(H) Let $\mathbb H \mathbb P^n$ denote the quaternionic projective space.  The group $SO(3)$ acts on $\mathbb H$ (by automorphisms of $\mathbb R$-algebras) by acting on the imaginary part in the obvious way, which gives rise to an action of $SO(3)$ on $\mathbb H \mathbb P^n$.
What is the quotient $\mathbb H \mathbb P^n/SO(3)$? I understand it's unlikely to be a manifold since the action fixes the real locus, but is there some nice description?
 A: For concreteness, I'm going to view $\mathbb{H}P^n$ as consisting of the points of the form $[z_0:...:z_n]$ with $0\neq (z_0,..,z_n)\in \mathbb{H}^{n+1}$, where $[z_0:...:z_n]$ is identified with $[z_0 q: .... z_n q]$ for any non-zero quaternion $q$.
Given an element $A\in SO(3)$, pick a unit length quaternion $p$ for which $\pi(p) = A$, where $\pi:Sp(1)\rightarrow SO(3)$ is the usual double covering map.  That is, under the identification of $\operatorname{Im}(\mathbb{H})$ with $\mathbb{R}^3$, conjugation by $p$ corresponds to left multiplication by $A$.  Note that conjugation by $p$ fixes the real component of any quaternion.
Then one way to write your action is as follows.  Given $A\in SO(3)$ and $z:=[z_0:...:z_n]\in \mathbb{H}P^n$, define $A\ast z = [p z_0:...:p z_n]$.  To see this is equivalent to your action, simply note that $[p z_0:...: pz_n] = [p z_0 p^{-1}:... pz_n p^{-1}]$, and that if we write $z_0 = r_0 + s_0$ as a sum of real an imaginary parts, $pz_0 p^{-1} = r_0 +As_0$.
As you noted, this action has fixed points:  it fixes precisely the subset $\mathbb{R}P^n\subseteq \mathbb{H}P^n$ given by the points containing a representative with all real coordinates.
In general, I don't of a nice way to representing the quotient.  If you're familiar with Lie groups, you can write it as $\Delta Sp(1)\backslash Sp(n+1)/Sp(n)\times Sp(1)$ where $\Delta Sp(1)$ is embedded into $Sp(n+1)$ as the set of multiples of the identity, and where $Sp(n)\times Sp(1)$ is embedded in the block form.  I'm willing to bet that for large $n$, this is the nicest description you'll find.
However, when $n=1$, the quotient $\mathbb{H}P^1/SO(3)$ is actually nice: it's homeomorphic to a closed disk $D^2$.  Here's one way to see that.
If we put the standard Fubini-Study metric on $\mathbb{H}P^1$, under the usual diffeomorphism $\mathbb{H}P^1\cong S^4$, this corresponds to the usual round metric.  The $SO(3)$ action is isometric, so, when viewed as an action on $S^4$, it must be linear.
There aren't many linear actions of $SO(3)$ on $S^4$.  In fact, as these corresponding to conjugacy classes of embeddings $SO(3)\subseteq SO(5)$, representation theory tells us there are precisely two.  The first is the usual block embedding $A\mapsto \operatorname{diag}(A,1,1)$, while the other is a bit stranger.  The important thing about the stranger one is that it has no fixed points.  (I can go into a lot more detail about this paragraph if you want).
So, the $SO(3)$ action on $S^4$ is the usual one.  One can view this as follows:  Start with the usual action of $SO(3)$ on $S^2$ and then suspend the action.  That is, define an action of $SO(3)$ on $\Sigma S^2$ by just copying the action on each "height".  Repeating this, we find the action of $SO(3)$ on $S^4$ is the same as the  $SO(3)$ action on $\Sigma^2 S^2$.  The action commutes with suspension, so the quotient $(\Sigma^2 S^2)/SO(3)$ is homeomorphic to $\Sigma^2( S^2/SO(3)) = \Sigma^2(point) = D^2$.
