So, I will work out the answer for a rotating ring of charge of radius $R$, which can be described by current density:
$$
\mathbf{J}=I\int_0^{2\pi} R.d\phi\,\left(\begin{array}\\-\sin\phi\\\cos\phi\\0\end{array}\right) \delta^{(3)}\left(\mathbf{r}-R\left(\begin{array}\\\cos\phi\\\sin\phi\\0\end{array}\right)\right)
$$
Where $I$ is the current in the ring. This is for an inertial observer and is in Cartesian coordinates. This can be fed into Biot-Savart law to find the magnetic field ($\mu_0$ is vacuum permeability):
$$
\mathbf{B}\left(\mathbf{r}\right)=\frac{\mu_0}{4\pi}\int d^3 r' \frac{\mathbf{J}\left(\mathbf{r'}\right)\times\left(\mathbf{r}-\mathbf{r}'\right)}{\left|\mathbf{r}-\mathbf{r}'\right|^3}
$$
For an observer (A) sitting on the XY-plane, at distance $\rho$ away from the origin, the magnetic field will be z-polarized:
$$
B_z\left(\rho\right)=\frac{I\mu_0}{4\pi}\int_0^{2\pi}Rd\phi.\frac{R-\rho\cos\phi}{\left(R^2+\rho^2-2R\rho\cos\phi\right)^{3/2}}=\frac{I\mu_0}{4\pi R}K\left(\rho,\,R\right)
$$
Where
$$
K\left(\rho,\,R\right)=\int_0^{2\pi}d\phi.\frac{R^2\left(R-\rho\cos\phi\right)}{\left(R^2+\rho^2-2R\rho\cos\phi\right)^{3/2}}
$$
is simply a short-hand for the unitless integral which I don't know how to evaluate (other than in the limit or numerically).
Let the full charge of the loop be $Q$, let the angular velocity of rotation be $\omega$. So $I=Q\omega/2\pi$ and:
$$
B_z\left(\rho\right)=\frac{Q\omega\mu_0}{8\pi^2 R}K\left(\rho,\,R\right)
$$
Next the electric field from the ring of charge density:
$$
\rho=\frac{Q}{2\pi}\int^{2\pi}_0 d\phi \,\delta^{(3)}\left(\mathbf{r}-R\left(\begin{array}\\\cos\phi\\\sin\phi\\0\end{array}\right)\right)
$$
For observer $A$ the electric field will be purely radial ($\epsilon_0$ is vacuum permittivity):
$$
E_{\rho}\left(\rho\right)=\frac{Q/2\pi}{4\pi\epsilon_0}\int_0^{2\pi}d\phi. \frac{\rho-R\cos\phi}{\left(R^2+\rho^2-2R\rho\cos\phi\right)^{3/2}}
$$
again, introducing:
$$
S\left(\rho,R\right)=\int_0^{2\pi}d\phi. \frac{R^2\left(\rho-R\cos\phi\right)}{\left(R^2+\rho^2-2R\rho\cos\phi\right)^{3/2}}
$$
the electric field becomes
$$
E_{\rho}\left(\rho\right)=\frac{Q}{8\pi^2 R^2\epsilon_0}\,S\left(\rho,R\right)
$$
Now consider:
$$
\frac{B_z}{E_\rho/c}=\frac{\omega R}{c}\,\times\frac{K\left(\rho,R\right)}{S\left(\rho,R\right)}
$$
Where $c$ is the speed of light. Playing with $r\approx R$ shows that for $K/S\sim 1$, so the key quantity here is $\omega R/c$ which is the ratio of the speed with which the charges move in the loop to the speed of light.
Clearly $\omega R/c<1$, but for large angular velocities it is, in principle, possible to have $c B_z/E_\rho>1$. Then all observers going through that point in space will observe non-zero magnetic field. The logic is based on $B^2-E^2/c^2$ invariant of the electromagnetic field
Under normal settings the electric and magnetic fields seen by the inertial observer will be, assuming the observer is displaced from the origin along the x-axis:
$$
\begin{align}
E_x &= E \\
B_z &\approx \frac{\omega R}{c} E
\end{align}
$$
From here you should be able to apply Lorentz tranforms to approximate the electromagnetic field in the instantaneous inertial frame of the rotating observer B
Noticed that ratio $K/S$ diverges for $r \uparrow R$, so if the observer is sitting on the loop you could probably have a situation $c B_z/E_\rho>1$ even for relatively mild rotation speeds
