Can any higher-dimensional Spheres be rotated everywhere equally?

You can rotate a circle so that every point on it (just the perimeter, not the interior) moves "equally". That is, every point moves with the same speed and even has the same "acceleration" (first-order derivative of velocity, which is the first-order derivative of a point's movement wrt time).

However, there is no way to rotate a sphere with this same character: the farthest points from the axis of rotation (equator) move much faster and have greater acceleration than the points closest to the axis (poles).

AFAIK, there is no closed (finite, no edges) 3-dimensional surface with this property. You can do it with a cylinder, but that is not a closed shape.

You could also do it with a torus by rotating it through the center (rather than around the center) the way a smoke ring does, but that's not a true "rigid" rotation (all points maintain the same distance to all other points throughout the rotation). The points on the inner part of the ring are closer together than those on the outside.

What I've been wondering is how this plays out in higher dimensions? Can the volume-surface of a 4D sphere be rigidly rotated equally like a 2D circle? Or does it fall to the same problems as 3D spheres? What about even higher dimensional spheres?

For that matter, are there any higher dimensional closed shapes that can be so rotated?

For clarity, I am only looking for "smooth" shapes, no significant discontinuities, and not just a disconnected set of points.

One way to make sense of what you wrote is: one can rotate odd-dimensional spheres without any fixed points.

A way to do this is: if you identify $\mathbb R^{2n}$ with $\mathbb C^n$, then at time $t$ you multiply by $\exp(2\pi i t)$.

On the other hand, this is impossible for every even-dimensional sphere, due to the hairy ball theorem.