Essentially, your questions boil down to the study of velocity fields arising from rigid motion. The short answer is that this is a well studied subject, the mathematical concept is that of the Lie algebra of the $3$-D special euclidean group. In general, you don't have a center of rotation due to possible translations, but if your motion is not purely translational, you always have an instantaneous axis of rotation.
Mathematically, writing $\vec v(\vec r)$ the velocity field at a given time, at position $\vec r$ with an arbitrary origin, you can find at all time, spatially constant $\vec v_0,\vec \omega$ such as
$$
\vec v = \vec v_0+\vec \omega \times r
$$
The motion is purely translational when $\vec omega = 0$. When it is not, there is always an instantaneous axis of ration is where $\vec v ||\vec \omega$, i.e. the line $\frac{\vec \omega\times\vec v_0}{\omega^2}+\mathbb R \vec \omega$.
To prove the above fact, you need to study the flow under rigid motions: $\vec y(\vec r,t)$ with $\vec v = \partial_t \vec y$, $y(\vec r,0)= \vec r$. Rigid motions are affine, so you can write $\vec y = \vec u(t) +R(t)\vec r$, with $R$ a time dependent rotation and the initial conditions give $\vec u(0)=0,R(0)=id$.
You immediately recoginze $\vec v_0 = \frac{d}{dt}\vec u$. The tricky part is convincing yourself that you can find $\vec \omega$ such that:
$$
\frac{dR}{dt}_{|t=0} \vec r= \vec \omega \times \vec r
$$
By taking the derivative of the dot product preserving property, you easily find that $\frac{dR}{dt}_{|t=0}$ is an antisymmetric operator, and you need to invoke the fact that all antisymmetric operators can be written as a cross product by a vector (specific to 3D).
Hope this helps.
