Relevant equation for calculating the time rate of change of a vector between an inertial and non-inertial rotating frames of reference:
$$ \left(\frac{d\boldsymbol r}{dt}\right)_s = \left(\frac{d\boldsymbol r}{dt}\right)_r +\boldsymbol \omega\times\boldsymbol r$$
where $s$ denotes the space fixed frame and $r$ denotes the rotating frame.
That is the relevant equation for the instantaneous co-moving inertial frame. What about a non-comoving space frame? Generalizing this to a space frame in which the origin of the rotating frame is moving results in
$$ \left(\frac{d\boldsymbol r}{dt}\right)_s =
\left(\frac{d\boldsymbol r_0}{dt}\right)_s
+ \left(\frac{d(\boldsymbol r - \boldsymbol r_0)}{dt}\right)_r
+\boldsymbol \omega\times(\boldsymbol r - \boldsymbol r_0)$$
where $\boldsymbol r_0$ is the displacement vector from the origin of the space frame to the origin of the rotating frame.
Suppose you use some other point $\boldsymbol r_1$ that is fixed from the perspective of the rotation frame (i.e., $\left(\frac{d(\boldsymbol r_1-\boldsymbol r_0)}{dt}\right)_r \equiv 0$. Go through the math (an exercise I'll leave up to you) and you'll find that
$$ \left(\frac{d\boldsymbol r}{dt}\right)_s =
\left(\frac{d\boldsymbol r_1}{dt}\right)_s
+ \left(\frac{d(\boldsymbol r - \boldsymbol r_1)}{dt}\right)_r
+\boldsymbol \omega\times(\boldsymbol r - \boldsymbol r_1)$$
In other words, the angular velocity $\boldsymbol \omega$ is independent of the choice of origin.
Best Answer
In general, yes. If different parts of the body have different angular velocities (or different angular accelerations, which will result in different angular velocities), then you have particles moving relative to one another. Assuming an object is a rigid body means that all particles are fixed in place relative to the body. If the body is not rigid, then this allows for relative motion, and hence different angular velocities.
Of course, you could always take a non-rigid body and move it in such a way as to make all parts of it have the same angular velocity or angular acceleration. But this is a contrived case.