To clarify, a rotation $P$ in 4-D is defined to be an orthogonal matrix with positive determinant, that is, $P^TP=I$, $\det P=+1$.
For every such rotation, one can find two perpendicular rotation planes meeting only at the origin. So $P$ has a matrix $$\pmatrix{\cos\theta&-\sin\theta&0&0\\\sin\theta&\cos\theta&0&0\\0&0&\cos\phi&-\sin\phi\\0&0&\sin\phi&\cos\phi}$$
If by axis you mean non-zero fixed vectors $Px=x$, then there are no axes in general for 4-D rotations, just as there are no axes in 2-D.
So none of the suggestions 1-3 hold: there are two orthogonal planes, each rotating independently, and causing the volume of vectors in between them to rotate with them. The projections of a body onto the planes rotate with the planes.
Add: A rotating 3-D sphere does have an axis, and there is only one way to extend the rotation to 4-D, namely by fixing the fourth dimension; this is the case $\phi=0$ in the above matrix.
If you are rotating with the sphere and look up at the sky, and you have 4-D vision, then the sky would not look like a dome but a volume (a 3-sphere). There would be a whole plane (appearing like a great circle) of fixed 'polar' stars, and all other stars rotate about this plane along with the 'equator'.
If you find it hard to imagine stars rotating about a plane, think of the flatman A. Square: all his stars rotate about his circular Earth and it might be inconceivable for him to imagine a fixed polar star.
Add2: Represent the 3-sphere using three angles $(\theta,\phi,\psi)$. Here is a plot of three random stars as they move in this space:
In general they are seen as moving in a direction (the one you would have moved if you weren't stationary) but rotating about that direction. The two rotations could of course have different rates.
