If you ask around about magnetic fields, you will read seemingly-authoritative articles which say magnetism is a consequence of length contraction. This is widely taught and is repeated in answers such as this one which talks about the magnetic force between current-carrying wires. I'm sure we're all happy with this force, which you can see described in the picture from Rod Nave's hyperphysics:
However note the concentric magnetic field lines around the wire. We know that a charged particle such as an electron will circle around these field lines, as per this depiction courtesy of Chegg homework help:
In addition we know that a positron will circle the magnetic field lines the other way. If the electron path is a left handed helix, the positron path is a right-hand helix. Or a circle, as per these pictures of an electron beam in a uniform magnetic field.
The length-contraction explanation for two wires moving together sounds fairly plausible. However there doesn't seem to be any way that length contraction in that linear wire can result in the opposite circular motion of electrons and positrons. The so-called explanations I've found are woeful, little more than smoke and mirrors and a rabbit from a hat. Would anybody care to have a crack at explaining how this happens? There will be a 100-point bounty to the least-worst answer.
How can length contraction result in electron circular motion in a magnetic field, and the opposite circular motion for a positron?
Edit: for clarification, see this drawing:
Best Answer
Review of the idea we're talking about
So let's review the mechanism by which this idea works. (In my view it makes little sense to critique the idea if we don't have a detailed understanding of it.) The idea is that, you've got currents on wires, and those wires (I'll call them wire-1 and wire-2) consist of nucleons and electrons. All of the nuclei are at rest relative to each other. In the rest frame of these protons in wire-1, you have electrons with a linear charge density $-\lambda_1$ moving with velocity $v_1$ relative to these fixed protons, which must have the same charge density $+\lambda_1$ to keep the wire electrically neutral in this rest frame. Now, in this frame of reference, the preferred way to work out the force on wire-2 is with the $\vec B$ field, as the $\vec E$ field due to wire-1 is zero. We see that there is no force from wire-1 on the protons of wire-2, but there is a force on the electrons of wire-2. So that's what we're going to analyze.
Now we boost by $v_2$ to get into the rest frame of the electrons in wire-2. If the electrons in wire-1 were moving with e.g. the same velocity $v_1 = v_2$, the distance between those wire-1 electrons expands by a factor of $\gamma$ while the distance between protons contracts by a factor of $1/\gamma,$ leading to both an $\vec E$ field and a $\vec B$ field. However, in this reference frame, the electrons in wire-2 are not moving and therefore their Lorentz force is not going to feel the $\vec B$ field from wire-1, only the $\vec E$ field, due to the modified charge density $(\gamma - 1/\gamma)~\rho$. (And that property is a fundamental consequence even when $u \ne v:$ if I boost into the rest frame of a particle, it feels no magnetic Lorentz force.)
Now if we were looking at a length of wire-2 in the proton-rest frame, having length $d\ell$ in that frame, then the charge along that length $-\lambda_2~d\ell$ gets transformed to the same-charge-at-different-times in the comoving frame, and the simultaneity-shift doesn't matter to us because the $\vec E$ field is constant over time. The total change in momentum per unit of proper time on these charges is therefore given by the $\vec E$ field $(\gamma - 1/\gamma)~\lambda_1~\hat r~/(2\pi\epsilon_0~L)$ times the charge $-\lambda_2~d\ell;$ but since $\hat r$ is orthogonal to the motion when we transform back we simply have to account for time-dilation, $dt=\gamma~d\tau$, and hence divide by one factor of $\gamma$. The result is a force $-(1 - \gamma^{-2})~\lambda_1~\lambda_2~d\ell~\hat r/(2\pi\epsilon_0~L)$ as transformed back into the reference frame of the protons.
And, as I'm sure you know, this matches up completely because $(1 - \gamma^{-2}) = v^2/c^2$ and $c^2 \epsilon_0 = 1/\mu_0,$ yielding $\mu_0~I_1~I_2~d\ell/(2\pi~L),$ just as the magnetic calculation works.
So, this is the mechanism by which "length contraction" in wire-1 leads to the force on wire-2: it is not strictly only "length contraction" at work here; the distance between electrons for example expands and there is a time dilation factor: but the core idea is that we can always boost into the reference frame of a particle, and then the magnetic field on that particle in that frame generates no Lorentz force, so the magnetic effects have all been pushed into the electric field, often with various length-contraction effects generating the newfound electric field.
Your actual question
You're (hopefully!) gonna kick yourself. It's just because the electric-only Lorentz force is $q \vec E$ and therefore is opposite if the charge of the particle is opposite. If you replace those electrons and nucleons in wire-2 with positrons and antinucleons, then the force on the antiprotons is easily seen to be zero; then we boost into the reference frame of the positrons, and we see the exact same electric field due to wire-1, but the Lorentz force on the positrons points the opposite direction (repulsive) because like charges repel and opposite charges attract.
In general it has to be that way: whenever you perform this "boost into its frame to make the $\vec B$ field have no Lorentz force" trick on any electron, you find some $\vec E$ field which provides the equivalent force; if you replace the electron with a positron with the same velocity, necessarily the fields calculated from these "length-contracted currents" is the same, but the force is opposite, because the positron has opposite charge from the electron.
Implications of solenoids.
You've asked this question in a much broader context, which involves a particle spiraling in a magnetic field. Well, if we want a constant magnetic field then the best way to get that is the inside of a solenoid. We'll draw this in two dimensions as a counterclockwise circle of (positive) current with magnetic field pointing "up" through the paper, constant inside the circle. We'll note directions on the paper with a compass rose: your charge $q$ is travelling "North" on the paper.
We boost into its reference frame. Again, the protons get uniformly length-contracted on both the East and West directions, increasing their density. The electrons in the East are travelling Southwards (because the counterclockwise current density is positive, the electrons go clockwise) and thus get doubly-length-contracted, increasing their density even more than the protons' density was increased.. So the eastern part of the solenoid has a net negative charge. The electrons in the West are travelling Northwards, in the direction of the motion, and thus experience length-expansion and decrease their density, so the western part of the solenoid has a net positive charge. This means that the $\vec E$ field in this setup is pointing from West to East, and a positron test charge will veer along with it, to the East, while an electron test charge will veer against $\vec E$ to the West. Rotational symmetry then guarantees that after a proper time $d\tau$, the situation that the particle sees is going to be exactly the same: if we relabel its present direction as "North" then it still sees a net positive charge in the "West" and a net negative charge in the "East" and veers similarly, so its net motion must be circular if we neglect the complicated radiation effects that come from having the particle accelerate in its rest frame.
So the positron orbits contrary to the (positive) current in the solenoid, and the electron orbits along with it, in just the way that you'd expect if you work out the magnetic field and the Lorentz force due to it.
Pretty picture
Edit: because you expressed some doubt that there is a charge accumulation due to the helicity of the solenoid electrons, I actually wrote a quick Excel sheet which solved for the $r^0=0$ hyperplane of the Lorentz boost of the helices $[ct,\;r\cos(\omega~t+\phi),\;r\sin(\omega~t+\phi),\;\mu~s + \nu~t]$ versus $[ct, r\cos(\phi),\;r\sin(\phi),\;\mu~s]$. As you can see, the green charges (positive nuclei) appear right of the blue charges (electrons) for most of the curves, indicating a higher charge density on the left versus that on the right. So yes, this is a real and direct prediction of the Lorentz transform and no, you cannot shake it. Moreover, I am claiming that this is the effect which generates the force in the reference frame co-moving with the particle: the particle cannot feel the magnetic part of the electromagnetic force tensor in that reference frame and instead feels a "sideways" force purely due to the electric part, which can be analyzed classically as due to charge accumulation on one side of the ring of current.