If that's the case, then please explain this.
Let's say that there is a test charge ##q## whose position from the origin is given by the radial vector ##\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} ## and there is a source charge ##Q## at the origin moving to the right along the positive ##x##-axis with a constant velocity ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over v} _s}## with respect to ##q##.
Now, in the rest frame of ##q##, ##q## has a velocity of magnitude ##0##, so it will not feel any magnetic effects. The force it will feel will be solely from the electric field generated by ##Q##. The force in the rest frame of ##q## is then:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = {{Qq} \over {4\pi {\varepsilon _0}}}{{\hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}$$
If we consider a difference reference frame in which we are moving to the right with a velocity of ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _o}##, then ##q## will have a velocity of ## - {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _o}## and ##Q## will have a velocity of ##{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _s} - {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} _o}##. The magnetic field then produced by ##Q## is:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
B} = {{{\mu _0}Q} \over {4\pi }}{{\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o}} \right) \times \hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}$$
The force on ##q## due to the magnetic field will then be:
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = - q\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o} \times \mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
B} $$
$$\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
F} = - {{{\mu _0}Qq} \over {4\pi {{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o} \times \left( {\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o}} \right) \times \hat r} \right)$$
So the total force the particle will experience in the moving reference frame is:
$${{Qq} \over {4\pi {\varepsilon _0}}}{{\hat r} \over {{{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}} - {{{\mu _0}Qq} \over {4\pi {{\left\| {\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
r} } \right\|}^2}}}\,{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o} \times \left( {\left( {{{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_s} - {{\mathord{\buildrel{\lower3pt\hbox{$\scriptscriptstyle\rightharpoonup$}}\over
v} }_o}} \right) \times \hat r} \right)$$
Which is clearly not equal to the force found in the stationary reference frame. Am I missing something here?