There's a theorem in Algebraic Topology that says any continuous map $f \colon S^{2n} \rightarrow S^{2n}$ has either fixed points or sends a point to its antipode.
Let $f$ be the function described as follows: Rotate the sphere away from you by pushing away the top side $90^{\circ}$, then rotate the sphere counter-clockwise $90^{\circ}$ by placing your hands on the left and right side of the sphere. Intuitively, from a standing position lie face down and then roll on to your left side.
What point is either fixed or sent to its antipode here? I can't see there being any fixed points as ever points seems to physically move, but I'm having trouble seeing what point specifically is sent to its antipode.
It is true that every point will move during this process, but two points will ultimately end up where they started.
In particular, for this combination of transformations, assuming you have the unit sphere centered at the origin, with the positive $x$-axis to the right, the positive $y$-axis away from you and the positive $z$-axis upwards, the point at $\left(-\frac{\sqrt3}3, -\frac{\sqrt3}3, \frac{\sqrt3}3\right)$ first gets sent to $\left(-\frac{\sqrt3}3, \frac{\sqrt3}3, \frac{\sqrt3}3\right)$ by the "away" rotation, then back to $\left(-\frac{\sqrt3}3, -\frac{\sqrt3}3, \frac{\sqrt3}3\right)$ by the counterclockwise rotation. Its antipodal point goes through a similar transformation.