Isoclinic rotations in four dimensions Given any collection of complementary, oriented (2D) planes in n-dimensional space, and an angle associated with each one, there is a unique rotation of the whole space which restricts to rotations in each plane by their given angles. Conversely, it's not impossible to prove all rotations of space arise in this way. Which leads to the question: can a single rotation arise in this way but from two different sets of complementary planes? If the attached angles are all distinct, I think the answer is no, which means the collection of stable planes is an invariant of "almost all" rotations (in the sense that the set of exceptions are of positive codimension in the group of all rotations).
If any two of the attached angles are the same, though, then there are an infinite number of planes which are stable under the rotation. Indeed, it is easy to verify that such a rotation moves every ray from the origin by the same angle. (Compare with rotation around an axis in 3D: if a ray is perpendicular to the axis then it is rotated by the full angle, but otherwise if the ray is closer to the axis, the before and after rays will be less than the full angle apart from each other.)
For simplicity, lets just take n=4 so our isoclinic rotations are rotations in two orthogonal planes by the same angle. Depending on if combining the orientations of the planes agrees with the orientation of the whole space or not, we can call them left isoclinic or right isoclinic rotations, because they correspond to multiplying quaternions (which form a 4-dimensional space) either on the left or on the right by unit quaternions. As such, the subgroups of left and right isoclinic rotations are both copies of S^3 within SO(4).
Now my curiosity is in which planes are stable under an isoclinic rotation, or without loss of generality a one-parameter subgroup of them. Any 2D plane has a unique orthogonal complement, and to the two resulting planes we get a one-parameter subgroup, which can be identified with a line in the corresponding lie algbera so(3) of dimension 3. Therefore we have some map from the Grassmanian manifold of 2D planes to the projective plane. Characterizing stable planes of a given one-parameter group of isoclinic rotations is then equivalent to finding the fibers of this map, which should all be identical by symmetery. So we have some fiber bundle
$$F\to \mathrm{Gr}(2,4)\to\mathbb{P}^2 $$
Is there a good explicit description of the fibers? Or just its isomorphism type?
My initial thoughts: Given the fiber $F\subseteq\mathrm{Gr}(2,4)$ of a point in $\mathbb{P}^2$ corresponding to some one-parameter subgroup $H$ of isoclinic rotations, if we have any point $x\in\mathbb{S}^3\subset\mathbb{R}^4$ the orbit $Hx$ is a circle in $\mathbb{S}^3$, which determines a plane $\pi$ in $\mathbb{R}^4$, and indeed $\pi\in F$. So we get a map $\mathbb{S}^3\to F$, and the fibers of this map should be the circles in $\mathbb{S}^3$ which intersect planes $\pi\in F$. Thus we have a fiber bundle $\mathbb{S}^1\to\mathbb{S}^3\to F$, which naively suggests to me $F\cong\mathbb{S}^2$ based on the Hopf fibration.
Some folks in chat though suggested $F$ should be $\mathbb{P}^2$ or a couple disjoint copies of it. Also mentioned was the fact that $\widetilde{\mathrm{Gr}}(2,4)$, the oriented Grassmanian which is a double cover of the usual $\mathrm{Gr}(2,4)$, is $\cong \mathbb{S}^2\times\mathbb{S}^2$, although I don't know why or for sure how to use this.
Ideas?
 A: Here is what I suspect is going on. Let $V$ be a $4$-dimensional oriented real inner product space. Starting from a $2$-dimensional subspace $W \to V$ of $V$, taking the exterior square produces a map
$$\Lambda^2(W) \to \Lambda^2(V)$$
and hence a $1$-dimensional subspace of $\Lambda^2(V)$. Now, the orientation and inner product allows us to define a Hodge star operation $\star : \Lambda^2(V) \to \Lambda^2(V)$, and the eigenspace decomposition of this operation splits $\Lambda^2(V)$ into two $3$-dimensional subspaces $\Lambda^2(V)_{+}, \Lambda^2(V)_{-}$, the subspaces of self-dual and anti-self-dual $2$-forms. This produces two maps
$$\Lambda^2(W) \to \Lambda^2(V)_{+}, \Lambda^2(W) \to \Lambda^2(V)_{-}$$
which I believe are both nontrivial; that is, we get two $1$-dimensional subspaces of two $3$-dimensional vector spaces. All of this induces an $SO(4)$-equivariant map
$$\text{Gr}_{2, 4}(\mathbb{R}) \to \mathbb{RP}^2 \times \mathbb{RP}^2$$
which is a double cover, and I believe the map you describe is one of the two components of this map, although I'm not sure which one. From here it shouldn't be hard to see geometrically that the fiber is $S^2$ (and that this $S^2$ is the same $S^2$ that appears as one of the two factors of the oriented Grassmannian) although I'm too lazy to work out the final details. 
A: $\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$Consider a consistently-oriented pair of planes, regarded as complex lines. In fact, you may as well fix an identification of $\Reals^{4}$ with $\Cpx^{2}$ and take the complex coordinate axes as $2$-planes. A left isoclinic rotation in these planes is precisely multiplication by a complex scalar, so it stabilizes every complex line through the origin, i.e., an $S^{2}$.
Alternatively, the standard complex structure $J$ is the left isoclinic rotation by $\pi/2$, and commutes with every left isoclinic rotation about the complex coordinate axes.
Particularly, this picture makes apparent that a left isoclinic rotation by some angle $\theta$, namely scalar multiplication by $e^{i\theta}$, is the same for every pair of orthogonal complex lines.
(Incidentally, rotation by a half-turn (a.k.a. $-I$) obviously stabilizes every plane through the origin, and this is the only non-trivial rotation to stabilize arbitrary $2$-planes. I'm guessing this was too trivial to mention.:)
A: The fiber is $S^2$.
Fair warning: this answer provides a correct solution to the question "What is the fiber of the given map", based on my tastes. It is not at all in the flavor of the question itself (which this answer mostly ignores), and thus I don't consider it a good answer to the question, which deserves some honest geometry. I'm posting this for my amusement and that of any passers-by.
1) The fundamental group of $\text{Gr}(2,4)$ is $\Bbb Z/2$.
2) Pulling back the $F$-bundle along the map $p: S^2 \to \Bbb{RP}^2$ induces a double cover of the total space $E \to \text{Gr}(2,4)$. By (1) this implies that $E = \widetilde{\text{Gr}}(2,4) \cong S^2 \times S^2.$ So we have a fiber bundle $F \to S^2 \times S^2 \to S^2$. Take the long exact sequence of homotopy groups to see that $\pi_1(S^2) \to \pi_0(F) \to \pi_0(S^2 \times S^2)$ is exact, and hence $F$ is connected; and that $\pi_2(S^2) \to \pi_1(F) \to \pi_1(S^2 \times S^2)$ is exact, so $\pi_1(F)$ is abelian and $F$ must be the projective plane or sphere.
3) A theorem due to Earle and Eells implies that $SO(3) = \text{Isom}(\Bbb{RP}^2) \hookrightarrow \text{Diff}(\Bbb{RP}^2)$ is a homotopy equivalence. Hence by the clutching construction projective plane bundles are classified by $\pi_1(SO(3)) = \Bbb Z/2$. (They're also all projectivizations of 3-plane bundles.)
4) Let's determine the two $\Bbb{RP}^2$-bundles over $S^2$. The trivial bundle's total space is $S^2 \times \Bbb{RP}^2$, which has nontrivial fundamental group . Following the clutching construction let's compute the fundamental group of the total space of the nontrivial $\Bbb{RP}^2$-bundle over $S^2$. Trivialize it over the top and bottom hemispheres. To apply van Kampen to these spaces, we need to understand the inclusions $S^1 \times \Bbb{RP}^2$ into the top half and bottom half of the bundle. We may as well assume we picked our trivialization so the inclusion into the "top" $D^2 \times \Bbb{RP}^2$ is the standard inclusion, so the induced map on fundamental groups is the projection $\Bbb Z \times \Bbb Z_2 \to \Bbb Z_2$.
The bottom inclusion (by the clutching construction) is $S^1 \times \Bbb{RP}^2 \to D^2 \times \Bbb{RP^2}$ is given by $(x, y) \mapsto (x,\text{rot}_x y)$, where $\text{rot}_x$ refers to the rotation by the angle $x$ in $SO(3)$. Following this rotation along the loop that generates $\pi_1(S^1)$ gives the trivial loop in $\Bbb{RP}^2$; so the induced map on fundamental groups is also the projection. Thus the fundamental group of the total space is $\Bbb Z/2$. This gives an answer to the question. (Actually, the universal cover of the total space of this bundle is the nontrivial $S^2$-bundle over $S^2$: $\Bbb{CP}^2 \# \overline{\Bbb{CP}^2}.$) One could also use Leray-Serre to show that the cohomology with $\Bbb Z/2$ coefficients of the total space is wrong.
5) So the total space of any $\Bbb{RP}^2$-bundle over $S^2$ has fundamental group $\Bbb Z/2$ and hence cannot be $S^2 \times S^2$. So the fiber of your fiber bundle must be $S^2$.

Given that $F$ fits into a fiber bundle $S^1 \to S^3 \to F$, like you claim, the homotopy long exact sequence immediately implies that $F$ is simply connected, proving the claim. But this is way less fun.
