Summary: Your friend is correct in that a rectangular wheel with their sides having ratio 2.95241:1 or larger, requires even more energy to push per unit length than a triangle, if after every push the wheel turns only a quarter turn (rectangular wheel) or third turn (equilateral triangle wheel), and comes to a full stop. A flat bar (one side essentially zero) requires the most energy to push per unit length.
For simplicity, let's assume the case where the shape only turns to the next vertex, and does not continuously roll. This means we can ignore the forward momentum, and only need to look at the potential energy changes.
(In practice, some of the potential energy is converted to forward kinetic energy, meaning only the initial push needs to overcome the potential energy barrier, and the following pushes only need to "top up" the potential/kinetic energy to keep the average speed constant. This is a bit too complex to go into detail here; consider asking this at physics.stackexchange.com to go "beast mode" into the actual details.)
Let $r$ be the minimum and $R$ the maximum radius of the wheel. (For regular polygons, these are the radii of the inscribed and circumscribed circles, with all three centers the same.)
At rest, the height of the axle from ground is $r$. To push the wheel to the next vertex, we need to push the center of weight over the vertex, and that means we need to push so that the axle lifts to $R$ from ground. After the tipping point, the wheel turns, but slow enough to stop at the next minimum energy state, turning all that potential energy into waste heat.
The change in potential energy in a near Earth gravity field is $\Delta E = m g h$, where $m$ is the mass, $g$ is the free fall acceleration, and $h$ is the change in altitude (distance from the center of Earth).
If we define our wheel so that it has perimeter 1, to push the wheel forwards 1 unit we need to do work
$$E = m g n (R - r)$$
where $n$ is the number of vertices in the wheel.
(For a circular wheel, $R = r$, so there is no potential energy change. Remember, we are ignoring the forward momentum, friction, and recovery of forward kinetic energy from potential energy; in this sense, pushing a circular wheel is "effortless", $E = 0$.)
For an equilateral triangle with side length $1/3$ (thus perimeter $1$), $R = 1/\sqrt{27}$ and $r = 1/\sqrt{108}$. Thus,
$$E_3 = m g \; 3 \left ( \frac{1}{\sqrt{27}} - \frac{1}{\sqrt{108}} \right ) = m g \frac{1}{\sqrt{12}} \approx 0.288675 m g$$
For a square wheel with side length $1/4$ (thus perimeter $1$), $R = 1/\sqrt{32}$ and $r = 1/8$,
$$E_4 = m g \; 4 \left ( \frac{1}{\sqrt{32}} - \frac{1}{8} \right ) = m g \left ( \frac{1}{\sqrt{2}} - \frac{1}{2} \right ) \approx 0.207107 m g$$
A pentagonal wheel has side length $1/5$ (thus perimeter $1$), $R = 2/\sqrt{250 - \sqrt{12500}} \approx 0.170130$, $r = 1/\sqrt{500 - \sqrt{200000}} \approx 0.137638$, and
$$E_5 = m g \; 5 \left ( \frac{2}{\sqrt{250 - \sqrt{12500}}} - \frac{1}{\sqrt{500 - \sqrt{200000}}} \right ) \approx 0.162460 m g$$
For a rectangular wheel with one side $w$, $0 \le w \le 1/2$, and the other side $1/2 - h$ (so perimeter is $2 h + 2 w = 1$), $R = \sqrt{w^2 + h^2}/4$, $r_1 = w/2$, $r_2 = h/2$, $E(0) = E_2$, i.e.
$$E_4(w) = \left ( \sqrt{ 8 w^2 - 4 w + 1 } - \frac{1}{2} \right ) m g$$
which reaches maximum at $w = 1/2$, $h = 0$ (or $w = 0$, $h = 1/2$), i.e. a flat bar as the wheel, with $$E_2 = m g \frac{1}{2} = 0.5 m g$$and minimum at $w = h = 1/4$, a square wheel, $E_4 = (1/sqrt(2) - 1/2 ) m g \approx 0.207105 m g$.
Interestingly enough, a rectangular wheel with sides $1/4 \pm \sqrt{\sqrt{3} - 1} / \sqrt{48}$ ($\approx 0.373495, 0.126505$) requires the same effort as an equilateral triangle wheel does. This means that a rectangular wheel with side ratio larger than $(1/4 + \sqrt{\sqrt{3} - 1}/\sqrt{48})$:$(1/4 - \sqrt{\sqrt{3} - 1}/\sqrt{48})$ requires more energy to push per unit length, than a equilateral triangle wheel does.
Best Answer
In the spirit of the no-words answer to the linked question: