The question in the title has to some extent been answered here: https://worldbuilding.stackexchange.com/questions/79619/does-earths-direction-of-rotation-affect-day-length?newreg=55b0054a895145b88d7572c29254a652 I find the answer a little unsatisfying, perhaps because I am approaching the problem like a mathematician (albeit flailing badly) rather than like a general education undergraduate taking astronomy as an elective. The issue I am having is that their is an assumption that the time deviation between a solar prograde rotation and the sidereal rotation is the same as the time deviation between the sidereal rotation and the (hypothetical) solar retrograde rotation.That's not completely obvious to me.
My mathematical recasting of this question is as follows: Suppose that we have a spherical planet orbiting a star in a perfect circle whose rotation is prograde and whose axis of rotation is not in the plane defined by the orbit. Let $r $ be the time (in seconds, say) of the prograde solar day and $s $ the time (in seconds) of the sidereal day. Suppose that the planet suddenly started spinning retrograde on the same axis with negative the original angular velocity. Let $t$ be the time (in seconds) of the retrograde solar day. Does it follow that $r-s=s-t$?
A handwritten solution that I wrote up for the original problem some years ago, possibly with assistance from the internet, defined $\alpha $ to be the angle of prograde rotation per prograde solar day; $\beta $ the angle of revolution per prograde solar day; and $\delta =\frac{\alpha }{\beta }$. Note that $\delta $ is the angle of prograde rotation per angle of revolution. By assumption, $-\delta $ is the angle of retrograde rotation per angle of revolution. I then assert that we wish to find $\gamma $ such that $$360 -\gamma \delta =\gamma .$$ I have no idea where that equation came from, and this is why I suspect that the argument was not original to me. A little later I state that $360-\gamma $ is the angle of retrograde rotation per retrograde solar day. With that assertion taken as a given, the rest of the original problem becomes straightforward.
I think it is simplest to start with sidereal days. Suppose there are $y$ sidereal days in a year, where $y$ is not necessarily an integer.
If the planet is rotating prograde, it then has $y - 1$ solar days per year.
If the planet were rotating retrograde, it would have $y + 1$ solar days per year.
For example, suppose there are $366.24219$ sidereal days in one year, which is $31556925.22$ seconds long. (This is based on a mean tropical year for Earth.) The sidereal day is therefore $86164.09$ seconds. Prograde rotation gives you $365.24219$ solar days of $86400$ seconds each, while retrograde rotation gives you $367.24219$ solar days of $85929.47$ seconds each. So prograde rotation gives you a solar day $235.91$ seconds (about $3$ minutes $56$ seconds) longer than the sidereal day, whereas retrograde rotation gives you a solar day $234.62$ seconds (about $3$ minutes $55$ seconds) shorter than the sidereal day.
So the worldbuilding answer was incorrect in saying that the solar day would be $3$ minutes $56$ seconds shorter than the sidereal day if the Earth rotated retrograde. But the error in this figure is less than two seconds.
It basically comes down to the fact that $$ \frac{T}{y} - \frac{T}{y+1} \neq \frac{T}{y-1} - \frac{T}{y} . $$ where $T$ is the length of a year in a fixed unit of time (such as seconds) and $T/y$ is the length of a sidereal day in the same units. The length of the solar day is \begin{align} \frac{T}{y-1} & \text{ with prograde rotation,} \\ \frac{T}{y+1} & \text{ with retrograde rotation.} \end{align}