Lets say the electrons are moving at v in the conductor frame and the ions are not moving. If you move at v, then the electrons are not moving in your frame, but the ions are moving at -v, so the current is the same.
If you move near the speed of light, then both electrons and ions are moving backwards in your frame of reference at close to the speed of light. It is necessary to transform the 4-current, which is
##J^\alpha = (c \rho, j)##
What is the four-current in the frame moving at v/2?
Assume total charge in this frame is 0.
The current is -nv/2 - nv/2
so the four-current is
##J_1 = (0, -nv)##
Let u = your velocity relative to the frame moving at v/2.
Using Lorentz transformation,
##J_2 = (\gamma uvn/c, \gamma nv)##
The current density is much greater in your frame, by a factor of ##\gamma##, and there is a charge density as well. But the length of the conductor will also be contracted by a factor of ##\gamma##, so the total current is the same. In this frame, there will be an electric field as well as a magnetic field.
You can look at it two ways. The electric field arises from a Lorentz transformation of the magnetic field in the original frame. Or the electric field arises from the charge which results from a Lorentz transformation of the current in the original frame. Same thing.