Step one is simplify. So we have a 1 dimensional lattice of equally spaced ($s$)positive and negative charges, labeled by position $n$:
...0...1...2...3...4...5
The position of the n-th positive ion vs. time is:
$$ x^+_n(t) = ns $$
The electrons all start moving simultaneously (in the lattice frame), and have positions:
$$ x^-_n(t) = ns + vt $$
The difference in position gives the linear charge densities:
$$ \Delta^+(t) = x^+_{n+1}(t) - x^+_n(t) = s $$
$$ \Delta^-(t) = x^-_{n+1}(t) - x^-_n(t) = \Delta^+(t) + (vt-vt)=s $$
So they are equal, and the wire is neutral.
Boosting to the moving frame ($v$) with the electrons:
$$ x'^+_n(t) = \gamma (ns -vt) $$
$$ x'^-_n(t) = \gamma (ns +vt -vt) = \gamma ns$$
Now get the right time coordinate:
$$ t = \gamma(t' + vx'/c^2) $$
so
$$ x'^-_n(t) = \gamma ns $$
$$ x'^+_n(t) = \gamma (ns -v\gamma(t' + vx'/c^2)) =\frac{ns}{\gamma}-vt'
$$
are the world lines in the electron frame.
The electrons are stationary (and further apart, which is shown in the Science Asylum video at 6:03), while the protons move to the left, and are contracted.
Here is a plot of 5 ions and 5 electrons moving to the right at $\gamma=2$:

Note that in the lab frame the densities are:
$$\rho_{\pm} = \pm 1$$
The magenta line is $t'=0$, aka the $x'$ axis, with uniform symbols at integer values of $x'$. It is easy to see that:
$$ \rho_{\pm} = \gamma^{\pm 1} = 2^{\pm 1} $$
Now I can boost that plot to the electron frame (so the magenta line becomes the abscissa):

That explicitly shows the electron density is reduced by $\gamma$, and the ion lattice contracted by $\gamma$.