Let the equations of motion be expressed in a frame with coordinates $q$. We now want to switch over to another (arbitrarily moving) frame, whose corresponding coordinates are $Q$, given by:
$$Q = f(q, t)$$
For example, if the frame itself is moving with position $x(t)$, we will have:
$$Q = q - x(t)$$
(where $x$ is not dynamic, but is completely specified in advance).
This is quite obviously, in the general case, just a point transformation that keeps changing with time; or, if you prefer, a different point transformation at different times. And that's the way one might expect it to be - this follows directly from the fact that the moving frame is moving.
This does not necessarily leave the equations of motion invariant. It's true that Euler-Lagrange equations (note that the Lagrangian must now be allowed to be time dependent)
$$\frac{d}{dt}\frac{\partial L(q, \dot{q}, t)}{\partial \dot{q}} = \frac{\partial L(q, \dot{q}, t)}{\partial q}$$
continue to hold, but the change in the form of the Lagrangian effected by the change of frame means that the equations of motion can 'look' different.
In the case that this point transformation is also a gauge transformation, we have a special situation. Consider the following/relevant example. In classical mechanics, from an inertial frame, the Lagrangian is:
$$L(q, \dot{q}) = \frac{1}{2}m\dot{q}^2 - V(q)$$
The general transformation to an arbitrary moving (non-rotating, for simplicity) frame is given by $q = Q + x(t)$, so that $\dot{q} = \dot{Q} + \dot{x}(t)$, and the Lagrangian becomes:
$$L(Q, \dot{Q},t) = \frac{1}{2}m\dot{Q}^2 + m\dot{Q}\dot{x}(t) + \frac{1}{2}m\dot{x}(t)^2 - V(Q+x(t)) $$
The term quadratic in $\dot{x}$ produces only a pure-boundary term in the action $S = \int L dt$, and is irrelevant. The main term of interest is the second one (responsible for the fictitious forces, and part of the generalized potential, as mentioned in this answer to the question you linked), and it's contribution to the action:
$$S_2 = \int m\dot{Q}\dot{x}(t)\ dt$$
Integrating by parts and neglecting the boundary term, we get
$$S_2' = -\int mQ\ddot{x}(t) \ dt$$
This readily gives the answer - for the "time-dependent point transformation", which corresponds to a change of frame, to be a gauge transformation, we must have $\ddot{x}(t) = 0$ for all time. In this case, we get the important part of the action as:
$$S' = \int \left[ \frac{1}{2}m\dot{Q}^2 - V(Q, t) \right] dt$$
This is hardly any different from the one we started with (the time dependence in $V$ is not an issue, and is just a reflection of the fact that the "field" would also appear to move in a moving frame; the important thing is that at a given time $t$, the particle sees the same force $-\nabla V$ at its location in both frames).
This is indeed, what makes inertial frames ($\ddot{x} = 0$ as seen from another inertial frame) special - the general point-transformation due to switching frames reduces to a gauge transformation, and the equation of motion 'looks' the same i.e. 'Galilean invariance'. That this doesn't occur in non-inertial frames leads to the fictitious forces seen in such frames.
Best Answer
just rotate the position vector to the mass.
$$\vec \rho= \left[ \begin {array}{ccc} \cos \left( \omega\,t \right) &-\sin \left( \omega\,t \right) &0\\ \sin \left( \omega\,t \right) &\cos \left( \omega\,t \right) &0\\ 0&0&1 \end {array} \right] \,\begin{bmatrix} x -R\\ y \\ 0 \\ \end{bmatrix}$$
I assumed that the rotation is about the z axes .
the kinetic energy is now
$$T=\frac m2 \vec{\dot{\rho}}\cdot\vec{\dot\rho}\\\\ \begin{align*} &\vec{\dot{\rho}}= \left[ \begin {array}{c} -\cos \left( \omega\,t \right) {\it \dot{x}}+ \left( \left( R+x \right) \omega-{\it \dot{y}} \right) \sin \left( \omega \,t \right) -\cos \left( \omega\,t \right) \omega\,y \\ -\sin \left( \omega\,t \right) {\it \dot{x}}-\sin \left( \omega\,t \right) \omega\,y+ \left( \left( -R-x \right) \omega+{\it \dot{y}} \right) \cos \left( \omega\,t \right) \\ 0\end {array} \right] \end{align*} $$
and you have one holonomic constraint equation
$$R^2=x^2+y^2$$