But the row of moving electrons is contracted in the lab frame, compared to what the cat sees. You can see that in your screencaps from the video too -- Derek sees 10 electrons per image width, whereas in the rest frame of the electrons there are only about 8½ electrons per image width.
What is potentially confusing is that as far as the electrons themselves are aware (electrons are not "aware" of anything, but never mind), they are not that the same mutual distance when they are moving as when the wire carried no current.
In other words, the row of electrons is not a rigid object. If each pair of neighboring electrons had been separated by a little rigid rod, the electrons would have to come closer together when the current starts flowing. But there are no such rods, and the row of electrons is free to stretch when the current begins to flow, and this stretching is exactly canceled out by the length contraction, such that in the lab it looks like the distance between the moving electrons is the same as the distance between the stationary protons.
What does the cat see? When the wire didn't carry current, the electrons and protons moved backwards together at the same speed (and with the same contracted spacing as the cat sees the protons have during the entire experiment). Then, when the current starts to flow, the electrons in front of the cat begin coming to a halt (with respect to the cat) slightly before those behind it. So from the cat's point of view the row of electrons get significantly stretched.
Meanwhile, Derek will see all the electrons begin to move at the same time. The cat and Derek do not agree whether two electrons changed their velocity at the same time or not -- this is relativity of simultaneity and is mathematically necessary to make length contraction consistent.
This problem generates lots of confusion. As with most of special relativity problems, the root cause is mistreatment of the relativity of simultaneity. Another problem is forgetting that it is a thought experiment, hence we don't really need to worry about the fact that electron drift velocity in real conductors is absolutely non-relativistic$^1$.
The first point to get wrong is that once the current starts, and the (thought experiment) electron are moving to the right relativistically, that their spacing must Lorentz contract. This is false, as it it contradicted by the fact that the wire remains neutral. This is a stipulation of the problem, so there is no need to ask "why doesn't their spacing Lorentz contract?"--see Bell's Space Ship Paradox.
The moving electron spacing must match the stationary ion density, in the wire's rest frame for this to be true.
From there: just draw it in Minkowski space:
In the above figure, we see a line of stationary (blue) singly ionized atoms (heretofore: protons..it's a thought experiment after all), and matching red electrons moving along +x with $\gamma=2$. The past light cone is green, and the magenta axis shows the $x'$ axis of the moving electrons, along with their coordinate dilation:
$$ U' = U \times \sqrt{\frac{1+\beta^2}{1-\beta^2} }$$
In $S$, the wire rest frame, we have $E\propto \rho+ - |\rho_-|=0$ and $B\propto I = \rho_-\beta \equiv -\rho\beta$
Note that the graph has 1 charge per length unit, so $\rho=1$.
Now do what must be done to resolve confusion: The Lorentz Transformation:
Now the electrons are at rest (vertical red lines), but their density is not 1:
$$ \rho_-' = \rho/\gamma = \frac 1 2 $$
Again: See Bell's Spaceship Paradox.
The proton density has increased to
$$ \rho'_+ = \gamma\rho = 2 $$
Hence, in the electron frame there is now an electric field:
$$ E' \propto \rho'_+ - |\rho'_-| = \gamma - \frac 1 {\gamma} = \gamma\beta^2$$
so that:
$$ E'_{\perp} = \gamma(E_{\perp} + \beta \times B) = \gamma(0+\beta(\beta\rho)) = \gamma\beta^2I = \gamma\beta B$$
The protons now constitute a current $I' = \gamma(-\beta) = -\gamma\beta=\gamma I$ so that:
$$ B'_{\perp} = \gamma(B_{\perp} + \beta\times E) = \gamma B$$
Thereby satisfying the standard Lorentz transformation of electromagnetic field.
1 For realistic electron velocities, $\gamma\approx 1$, which seems like it should break the analysis (intuitively). The change in density is dominated by the relativity of simultaneity. If you imagine a time in the wire rest frame when each proton has an electron next to it (call that a series of coincident events $E_n(t_0, x_n)$ where $x_n = n \times [\rm lattice spacing]$ and then consider the times $t'_{0, n}$ in the moving frame, those events occur in the future (present) and ((past)) for $n<0$ ($n=0$) (($n>0$)), showing that the electron density is less than the proton density in the moving frame...and $\gamma$ his little influence.
Best Answer
Not all problems work out nicely in 3 + 1 dimensions. Consider two 4-currents in the (unprimed) wire rest frame:
$$ j^+_{\mu} = (\rho^+, 0, 0, 0) $$
$$ j^-_{\mu} = (\rho^-, v_d\rho^-, 0, 0) $$
With the charge densities satisfying $\rho^+ = -\rho^-$ (the wire is neutral) and the drift velocity is leftward, $|v_d| = -v_d$, the total four current density is:
$$ j_{\mu} = (0, v_d\rho, 0, 0) $$
(note: many complain the drift velocity is too small to invoke $\gamma$, which is true, but if you do the Lorentz transform parallel to the current you will see it is the relativity of simultaneity that dominates the change in charge density).
The moral of that parenthetical story is: when in doubt, Lorentz transform. So that's what we need to do here. With ($c=1$) and the test charge's (primed) frame moving along $+y$:
$$ u_{\mu} = \gamma(1, 0, v, 0) $$
you get:
$$ \Lambda_{\mu\nu} = \left [\begin{array}{cccc} \gamma & 0 &-v\gamma & 0 \\ 0 & 1 & 0 & 0 \\ -v\gamma & 0 & \gamma & 0 \\ 0 & 0 & 0 & 1 \end{array}\right ]$$
$$ j'_{\mu}= \Lambda_{\mu\nu}j_{\nu} $$ $$ j'_{\mu}= (0, v_d\rho, 0, 0) = j_{\mu} $$
So the force in the primed frame is caused by a magnetic field, no 3 + 1 magic here.
Of course, this is wholly unsatisfying compared with the original problem, which works out in spite on being not at all manifestly covariant.
It also raises the question of "where is the moving charge?". If you're moving above the wire, then the transformation of a magnetic field is:
$$ B'_{\parallel} = B'_{\parallel}$$
That is fine, but if you're moving towards the wire (in the plane of the current and charge):
$$ B'_{\perp} = \gamma\big(B_{\perp} - \vec v \times \vec E\big) =\gamma B_{\perp}$$
as $\vec E = 0$.
So now we've picked up a $\gamma$: the primed field is stronger (and not circular symmetric about the wire). That is not evident from the simple transformation of $j_{\mu}$.
I think the resolution here is to go to Jefimenko's Eq for a magnetic field generated by currents and charges on the past light cone:
$$\vec B'(\vec r', t') = -\frac{\mu_0}{4\pi}\int{\Big[ \frac{\vec r' - \vec r''}{|\vec r' - \vec r''|^3} \times \vec J'(\vec r'', t'_r) + \frac{\vec r' - \vec r''}{|\vec r' - \vec r''|^2} \times \frac 1 {c} \frac{\vec \partial J'(\vec r'', t_r)}{\partial t'} \Big]dV''}$$
Since the current is constant:
$$\vec B'(\vec r', t') = -\frac{\mu_0}{4\pi}\int{\Big[ \frac{\vec r' - \vec r''}{|\vec r' - \vec r''|^3} \times \vec J'(\vec r'', t'_r)\big]dV''}$$
(Note the double primed coordinate here is the integration variable, not any reference frame).
Computing this a bit of a project. In the primed frame you have a current density looking like:
$$\vec j'(x', y', z', t') = v_d Q\delta(y'-vt')\delta(z')\hat x' $$ where Q is the limit of $\rho$ in an infinitely thin wire. So this is a line of moving current.
Then fix ${\vec r'}$ to be the position of the charge, in the primed frame:
$$ \vec r' = (x', y', z') = (0, 0, 0) $$
At this point, it is best fix $t' \ne 0$ so you have some distance from the wire.
Then compute the retard time $t'_r$. The $\delta(y'-vt')$ will filter over the past light cone along the moving wire, giving some kind of conic section, but it should all work out, as it always does. Have at it. $\gamma$ is in there, somewhere.