This isn't really a solution - just rewriting it in terms of a fourth-order differential equation. I've also cross-posted this answer on MathOverflow.
Let $$V(t):=\pi\int_0^tr(s)^2ds$$
We can then rewrite $r(s)$ as $$r(s)=\sqrt{\frac{V'(s)}{\pi}}$$
If we rewrite $S_{ave}$ using just $V$, we get $$S_{ave}=\pi\sqrt{\frac{V'\left(0\right)}{\pi}}+\frac{\pi^{2}}{V\left(h\right)}\int_{0}^{h}\left(\left(\frac{V'\left(s\right)}{\pi}\right)^{2}+2\sqrt{\frac{V'\left(s\right)}{\pi}}\int_{s}^{h}\frac{V'\left(z\right)}{\pi}dz\sqrt{1+\frac{1}{4\pi}\cdot\frac{V''\left(s\right)^{2}}{V'\left(s\right)}}\right)ds$$
The inner integral simplifies to $\frac{1}{\pi}(V(h)-V(s))$, so this simplifies to (getting rid of some of the $\pi$ terms as well) $$\sqrt{\pi V'(0)}+\frac{1}{V(h)}\int_{0}^{h}\left(V'(s)^2+2\cdot(V(h)-V(s))\sqrt{\pi V'(s)+\frac{V''(s)^2}{4}}\right)ds$$
We want to minimize this value assuming a fixed $V(h)$. Assume we have a fixed $V'(0)$ and $h$ as well. We then want to minimize $\frac{\sqrt{\pi V'(0)}}{h}+\frac{1}{V(h)}\mathcal{L}(s,V, V', V'')$, where $\mathcal{L}(s,V, V', V'')$ is given by $$(V')^2+2\cdot(V_h-V)\sqrt{\pi V'+\frac{(V'')^2}{4}}$$
and $V_h=V(h)$. We can then use the Euler-Lagrange equation to get that the stationary points of the average surface area (with respect to $V(s)$) would be given by $$\frac{\partial\mathcal{L}}{\partial V}-\frac{d}{ds}\left(\frac{\partial \mathcal{L}}{\partial V'}\right)+\frac{d^2}{ds^2}\left(\frac{\partial \mathcal{L}}{\partial V''}\right)=0$$
This ends up being a fourth-order differential equation with a long form (I started writing it out before realizing that the last term would make it be very long).
Edit: Using the Beltrami identity, which TheSimpliFire mentioned in a comment, and this answer, we can write $$\mathcal{L}-V'\frac{\partial\mathcal{L}}{\partial V'}+V'\frac{d}{ds}\frac{\partial \mathcal{L}}{\partial V''}-V''\frac{\partial \mathcal{L}}{\partial V''}=C$$
Plugging in $\mathcal{L}$ and simplifying, we get $$-V'(s)^{2}\left(1+\frac{V''(s)\left(4\pi V'(s)+V''(s)^{2}\right)+4\pi\left(V(s)-V_{h}\right)\left(2\pi+V^{(3)}(s)\right)}{\left(4\pi V'(s)+V''(s)^{2}\right)^{\frac{3}{2}}}\right) = C$$
Best Answer
We need a clean parametrization of one of these twisted sides. To this end put ${\pi\over n}=:\alpha$. Then one edge of the polygon in its initial position is given by $$u\mapsto(r\cos\alpha, r\sin\alpha\> u,0)\qquad(-1\leq u\leq1)\ .$$ The twisting with angular velocity $\omega=1$ turns this edge counterclockwise and at the same time moves it in the vertical direction with a certain speed $\lambda$. It follows that we obtain a helical band with the parametric representation $${\bf r}(u,t)=\bigl(r\cos\alpha\cos t-r\sin\alpha\>u\,\sin t,\>r\cos\alpha\sin t+r\sin\alpha\>u\,\cos t,\lambda\, t\bigr)$$ whereby $-1\leq u\leq 1$ and $t\geq0$. One then has to compute $$d\pmb\omega(u,t)={\bf r}_u(u,t)\times{\bf r}_t(u,t)$$ and in the sequel the scalar surface element $${\rm d}\omega(u,t)=\bigl|d\pmb\omega(u,t)\bigr|\ .$$ This is the quantity that has to be integrated over $-1\leq u\leq1$ and $0\leq t\leq T$, whereby $T$ has to be chosen according to your specifications. The resulting integral is elementary; but in any case: The claim that the area of the side is equal to the untwisted area is wrong.