When we say that a current is made of moving charge carriers (like electrons), we mean that the charge carriers are moving relative to some frame. Often, that frame is "us", or more precisely, the lab.
However, all inertial observers must be able to derive the same consequences for electromagnetic phenomena using Maxwell's equations. A moving charge near a wire experiences a magnetic force, but we can surely go into the rest frame of the charge, in which case $v\times B=0$; how can we explain the deflection of the charge in this frame.
In fact, you can find such a frame, where the test charge's is zero. And in this frame, there is no magnetic force. However, while the negative electrons move in the lab frame, the positive ions making up the crystal structure of the wire do not. As a result, when moving into the frame where the current is zero, if one carefully accounts for length contraction, one finds that the electrons and ions contract by different amounts, leading to a net charge density in the wire. This creates an electric field around the wire. This electric field is then responsible for creating an electric force that produces the same physical effect on a test charge, that could also be observed by the test charge interacting with a magnetic field in the lab frame.
This reflects a general phenomena, that electric and magnetic fields themselves are not invariant under boosts -- meaning different inertial observers will disagree about how much electric field there is. However, their combined effect as an electromagnetic field is the same for all inertial observers. You can read more about that here: https://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity
In fact, trying to reconcile why different inertial observers would assign different "causes" to electromagnetic phenomena is what led Einstein to special relativity, which is why his famous paper on special relativity is called "On the electrodynamics of moving bodies".
Thanks @Dale for a useful correction to the the original version of this answer.