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 S^{3} fibration.
The Cartesian product of the complex plane with itself houses a unit 4D ball at the origin; its surface is S^{3} on which we define the great circles. We choose two complex numbers x and y which locate a point on S^{3} when xx* + yy* = 1. The distance of (x, y) from the origin, (0, 0) is √(xx* + yy*) and thus S^{3} = {(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 S^{3}, (w, x) and (y, z) are on the same plane just if wz = xy.
Such a plane intersects S^{3} 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 S^{2}. Fiber bundle wise S^{2} is the base and S^{1} 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 S^{3} 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 S^{3} 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 ℂ^{2} are symmetries of the fibration:
 maps each fiber onto itself.  
 rotates about the fiber {(cos θ, sin θ, 0, 0)  0≤θ<π} = {(e^{iθ}, 0)  0≤θ<π}.  

Consider the following map f from our S^{3} 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 S^{3} to SO(3) but it is a double cover: g(w) = g(−w).
The point (1, dx, dy, dz) is in S^{3}. 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.