This spherical geometry question involves isometries. I am particularly looking for isometries with no fixed points.
Here's a relatively straightforward one, built by composition:
First, flip your sphere about the $xy$-plane; this ensures that every point formerly in the northern hemisphere is now in the southern hemisphere, and vice versa — and importantly, it leaves points on the equator unchanged.
Next, rotate about the $z$ axis by, e.g., $\frac\pi4$; this maps the hemispheres to themselves (so that we can be certain that none of the points off the equator is a fixed point), and displaces every point on the equator (so that none of the points on the equator is a fixed point either).
Note that both coffeemath's example in comments and mine here are orientation-reversing. This isn't an accident; it can be shown that all orientation-preserving isometries of $S^2$ have a fixed point (and in fact, that all orientation-preserving isometries of $S^2$ are rotations about some axis) : see http://en.wikipedia.org/wiki/Euler%27s_rotation_theorem .
Contrariwise, this implies that all orientation-reversing isometries of $S^2$ can be found by composing the antipodal map $\rho$ with a rotation about some axis: since the composition of $\rho$ with the orientation-reversing isometry $\iota$ is an orientation-preserving isometry, $\rho\iota$ must be orientation-preserving, and therefore a rotation about some axis — and then $\iota=\rho^2\iota=\rho(\rho\iota)$ is the composition of $\rho$ with this rotation.
This also implies that that the only orientation-reversing isometries of the sphere that do have fixed points are the reflections about some plane through the origin: we can write the isometry as $\iota=\rho r$ for some rotation $r$ and consider the hemispheres and equator of the sphere with respect to the axis of the rotation $r$. Then just as in my example above, no point not on the equator of $r$ can be a fixed point of $\iota$ (because it winds up in a different hemisphere); similarly, since each point on the equator of $r$ is rotated and then mapped to its antipode, the only way in which any of these points can remain fixed is if the rotation $r$ is a rotation by $\pi$ — in which case $r$ itself maps every point on its equator to that point's antipode, and so the composition $\iota=\rho r$ is equivalent to reflecting all of $S^2$ about the equator of $r$, leaving that equator fixed in the process.