Would a spinning, evenly charged sphere generate a magnetic field? I could see how it would go either way.
On the one hand, there IS moving charge; on the other hand, it is not discernibly moving.
 A: What you call "discernibly" moving, is called a non-stationary current-density. Consider a wire with constant charge-flow at every cross section. Then this current is in the same sense not "discernibly" moving as your charge on the sphere. Still there is a magnetic field around the wire.
You can think of the charged sphere as infinitely many circular wires around the axis of rotation. All these wires create magnetic fields that add up. So yes, there is a magnetic field due to a rotating sphere.
For a calculation approach I suggest this: https://physics.stackexchange.com/a/173022/75518
A: The things that induce a magnetic field are a current or a changing electric field (with respect to time). In this situation you have charges moving,so you have a J.dS, which means you have a current.  
EDIT:
If you prefer a different approach, then consider the sphere to be an infinite amount of circular circuits (each with different radius because of the spherical surface).
So you know that when you have moving charges in a circuit, it means that you have current, and thus you have induced magnetic field.
So now you have infinite circular circuits one next to another. (Think of a solenoid with infinite turns. The same approach with different geometry can be made here-only for intuition though.)
A: You have a moving charge and a circular moving that is an acceleration for what you need a force. Remember the Lorentz force $ \vec F = q \vec v \times \vec B $. This vector cross product can be rewritten to $$ \vec B = \dfrac{\vec F \times \vec{qv}}{\|\vec{qv}\|^2}$$.
Edit: The equation was edited following https://math.stackexchange.com/questions/1219103/how-to-rewrite-a-vector-cross-product/1219127#1219127
