Suppose two electrons are moving along parallel paths at constant velocity. Then wouldn't the repulsion of the two electrons be slightly reduced because of the magnetic fields?
If so, how is the apparent decrease in force between the two particles explained in the frame in which both electrons are at rest?
The force between the electrons (like all forces in Special Relativity) is not a Lorentz invariant and so we would not expect it to be the same in all frames of reference.
Both electric and magnetic fields need to be transformed to their forms in any new frame of reference and the Lorentz force recalculated.
i.e. $$ q(\vec{E} + \vec{v}\times \vec{B}) \neq q(\vec{E}' + \vec{v}'\times \vec{B}')\ . $$
In this case, the relevant frames of reference are the rest frame of the electrons and the frame in which they are observed to be moving. The transformations are straightforward and you find that the repulsion between the electrons is maximised in the electron rest frame, but reduced by the Lorentz factor $(1- v^2/c^2)^{-1/2}$ in the other frame, where $v$ would be the electron speed.
There is no decrease of the electrostatic repulsion force (and magnetic field) between the two electrons in the co-moving reference frame where the electrons are at rest. The magnetic field appears only for a reference frame that is not moving with the electrons. This can be understood in the framework of Special Relativity which gives a simple transformation of electric and magnetic fields between inertial reference frames. See e.g., D.J. Griffiths, Introduction to Electrodynamics, Prentice-Hall 1999.
