The Hopf fibration partitions S3 into great circles. There is much on the Internet about this construction including pictures, which are necessarily somewhat incomplete. Penrose calls the construct a Clifford bundle and it is indeed an important fiber bundle.

The construct is more symmetric than indicated by images on the web. Some web sites speak of filling three space with circles. The 2D images on the net cannot show these symmetries. Even the depicted flat 3D configurations lack much of the symmetry of the S3 fibration.

The Cartesian product of the complex plane with itself houses a unit 4D ball at the origin; its surface is S3 on which we define the great circles. We choose two complex numbers x and y which locate a point on S3 when xx* + yy* = 1. The distance of (x, y) from the origin, (0, 0) is √(xx* + yy*) and thus S3 = {(x, y) | xx* + yy* = 1}. Choosing two complex numbers X and Y, not both zero, {(x, y) | Xx + Yy = 0} is a 2D plane thru the origin, and indeed any 2D plane thru the origin is of this form for some X and Y, not both zero. If we multiply X and Y by the same complex non zero value Z, the plane is unchanged. Two points in S3, (w, x) and (y, z) are on the same plane just if wz = xy.

Such a plane intersects S3 in one of our great circles and each of our great circles is such an intersection. The ratio of X and Y thus identifies the great circle but we must include the point at infinity among these ratios. The topology of these ratios is S2. Fiber bundle wise S2 is the base and S1 is the fiber.

Viewed as an equivalence relation where points on the same fiber are equivalent, a symmetry of a fibration is a transformation on the space that leaves equivalent points equivalent. Any point in S3 is like any other point, and then there is one additional degree of symmetry as we rotate about the fiber thru that point. The set of symmetries thus has 4 dimensions and is a fiber bundle whose base is S3 and whose fiber is a circle. I think that the entire symmetry group consists of 4 disconnected copies of the bundle described above. The following linear transformations on ℂ2 are symmetries of the fibration:
eiθ0
0eiθ
maps each fiber onto itself.
10
0eiθ
rotates about the fiber {(cos θ, sin θ, 0, 0) | 0≤θ<π} = {(eiθ, 0) | 0≤θ<π}.
01
10
This is a nonlinear symmetry: f((x, y)) = (x*, y*)
I think the above transformations generate all the symmetries.

Consider the following map f from our S3 to the quaternions as reified here:
f((a + bi, c + di)) = (a, (b, c, d)).
The function q2m maps a unit quaternion into an orthogonal matrix in SO(3) and that matrix is the transformation on 3D space for which quaternions were invented. The function g(w) = q2m(f(w)) maps from S3 to SO(3) but it is a double cover: g(w) = g(−w).

The point (1, dx, dy, dz) is in S3. Near that point the great circles pass thru in the dx direction. Indeed {(cos θ, sin θ, 0, 0) | 0≤θ<π} is a fiber.

Consider the tangent bundle of S2 and then the sub-bundle of those vectors in the tangent spaces with length 1. Penrose notes that this latter bundle is isomorphic to the Clifford bundle. It is well known in consumer topology that you can’t comb the hair on sphere and this leads to the conclusion that there are no cross sections of the Clifford bundle.