How can position of charge moving under magnetic field have different positions for different observers

Assume we have a point charge $$Q$$ moving with velocity $$\vec{v}$$ with respect to an observer at rest and is in the presence of magnetic field $$\vec{B}$$. $$\vec{B}$$ and $$\vec{v}$$ are perpendicular to each other.

For the observer at rest, the charge would start doing a circular motion in the plane which contains the velocity and magnetic field vectors.

But if an observer was moving with $$\vec{v}$$ , the charge wouldn't experience any force and remain at rest. The moving observer would keep traveling in a straight line with respect to a stationary observer. As the charge is at rest with respect to the moving observer, we can say that it should also be traveling in a straight line with respect to a stationary observer.

This seems to be a rationally impossible to me. Please explain this paradox.

Thank you

The magnetic field in the rest frame is (partially) transformed into an electric field in the frame initially co-moving with the charge. In this frame, the motion of the charge is initiated by the electric component $$Q\mathbf E$$ of the Lorentz force. Thereafter, the charge moves under the influence of a combination of electric and magnetic forces.

Electric and magnetic fields transform as part of a single rank 2 tensor in response to a change of coordinates. The Lorentz transformation is the change of coordinates between frames moving at constant velocity ("Lorentz boosted") with respect to each other, such as the two frames of interest here. The transformation equations can be found here. Just to make this answer self-contained, in vector form these are $$\mathbf E_\parallel'=E_\parallel$$ $$\mathbf B_\parallel'=B_\parallel$$ $$\mathbf E_\perp'=\gamma \left(\mathbf E_\perp + \mathbf v \times \mathbf B_\perp\right)$$ $$\mathbf B_\perp'=\gamma \left(\mathbf B_\perp - \frac{\mathbf v}{c^2} \times \mathbf E_\perp\right).$$ Here, the primed vectors are the fields in a frame moving with velocity $$\mathbf v$$ relative to the "unprimed frame", and $$\gamma = 1/\sqrt{1 - v^2/c^2}$$. $$\parallel$$ denotes the field components parallel to $$\mathbf v$$, and $$\perp$$ the components perpendicular to it.