(...) my question is, can spin be related to a rate of change of
anything at all with respect to time? Spin may not be related to
rotation in $\mathbb{R}_3$, but can we relate it to a rotation or other kind of
motion in some other space, possibly a non-Euclidean space?
It really depends on what you mean by "change of anything with respect to time", but the general answer is no, not in the way you are probably thinking of it.
No, because it is not correct to think of the spin angular momentum (SAM) of a particle as due to any kind of "rotation" or other "movement" in any (Hilbert) space.
An electron with spin $+1/2$ is not rotating, nor changing, for the sole fact of having a SAM.
In fact, if no external influences act on the electron, such spin state (or any other spin state for that matter) will be an eigenstate of the system, meaning that by definition it is stationary: nothing has to "move" for that electron to possess such SAM.
If that were the case, it would mean that spin is not really a "fundamental" quantum number, but only a "feature" of some more fundamental property.
It is important to notice that a similar argument also holds for orbital angular momentum: a particle (or composite particle, or any other type of quantum state) with a definite orbital angular momentum is not necessarily something that can be thought of as "rotating around": a single particle with an appropriately structured spatial wavefunction can have a definite orbital angular momentum, yet not be rotating in any sense of the word (on the other hand, it is true that the orbital angular momentum is generally a feature of the spatial profile of the wavefunction, and is therefore in this sense not a fundamental property of the system).
We like to picture the electrons in atoms as "rotating" around the nucleus, but that picture is really as wrong as the picture of a rotating spin: the electrons (or better said, the nucleus+electron systems) are in a stationary state, meaning that nothing is changing with respect to time.
The difference in such cases is however that the orbital angular momentum quantum number is completely written in the structure of the spatial wavefunction, and is therefore by no means an intrinsic property of the particles.
Yes, in the sense that of course the spin state can be changing with respect to time, or other properties of the particle be changing with respect to time depending on their spin state.
Put an electron in an appropriate magnetic field and you will see it move in one direction or the other depending on its spin state.
Or the spin direction itself can be rotating with a fixed angular frequency, like what happens to nuclei in an NMR machine.
But SAM seems so much like a regular angular momentum!
Does it?
Do you think of a single photon with a definite value of the polarization as rotating? Probably not, but you should! The "polarization" of light is nothing but a different name for the spin of photons.
Yet in the case of photons most people seem to not have so many difficulties in seeing the polarization as something unrelated to a kind of rotation.
Can the expectation value of the SAM operator be related to the rate of change of the expectation value of some "classically interpretable" operator?
No! The only way for this to make sense is the existence of a "classical equivalent" of SAM.
There is no such thing.
Why? Plainly said, because there isn't: classical objects are really just only characterized by position and momentum.
But what about Ehrenfest theorem?
What about it? Sure you can apply it, in its general form stating the relation between the variation of the expectation value of an operator and the expectation value of the commutator of the operator with the Hamiltonian:
$$
\frac{d}{dt}\langle A \rangle = \frac{1}{i\hbar}\langle [ A, H] \rangle,
$$
holding for any operator $A$ measuring an intrinsic (that is, time-independent) property of a particle.
If you take $A=S_z$, or some other component of the spin, you will get an expression for the rate of variation of the average value of the SAM in the particular system under consideration.
If it is an electron in an electromagnetic field you will get an expression for the time variation of $\langle S_z \rangle$ with respect to the particular electromagnetic field applied, which will tell how the spin of the particle evolves, but not much more.
If you instead want to find some operator $A$ such that
$$
\frac{d}{dt}\langle A \rangle \simeq \langle S_z \rangle,
$$
you can probably find it: you need to find a system described by an Hamiltonian $H$ such that there is an operator $B$ such that $[A,H] \simeq S_z$.
I see no reasons for this to not be possible, but it will hardly give you any insight into the nature of SAM, as it will be highly dependent on the particular system you are considering.
In fact, you shouldn't think of Ehrenfest theorem as something giving general connection between classical and quantum mechanics, as it is nothing but an alternative and equivalent way to state Schrödinger's equation.
But why is spin angular momentum so strictly related to the orbital angular momentum then?
Because the quantum object "corresponding" to the classical angular momentum is the total angular momentum operator, $\boldsymbol J = \boldsymbol L + \boldsymbol S$.
This is a consequence of the fact that the Lagrangian describing the interactions between the various particles only preserves the total angular momentum, not singularly the spin or orbital components of it.
This is not so unnatural as it may at first appear.
Think for example at the 3D wavefunction of a photon or electron.
While we are used to think of one such particle as having some value of SAM, in general you will have a probability amplitude of the particle having a value of SAM for every point of the wavefunction (or equivalently said, entanglement between SAM and position).
The SAM operator $\boldsymbol S$ acts on such wavefunction rotating the SAM degrees of freedom at every point in space, leaving unchanged the spatial amplitude distribution.
The orbital angular momentum operator $\boldsymbol L$ on the other hand rotates the spatial distribution of amplitudes, carrying around untouched the spin degree of freedom associated to each point.
Now, why should this whole complicated mess be invariant when you apply only one of these two operations?
Indeed in won't in the general case: you will need to rotate the spatial distribution and the internal degrees of freedom accordingly to obtain the same structure.