What's the height of the pyramid ?

Clearly in both cases the volume is the same.

Let $x$ be the length of a side of the $n$ gon.

$\dfrac{x}{2} = r\sin(\dfrac{\pi}{n}) \implies r = \dfrac{x}{2\sin(\dfrac{\pi}{n})} \implies$ $h = r\cos(\dfrac{\pi}{n}) = \dfrac{1}{2}x\cot(\dfrac{\pi}{n}) \implies$

The area $A$ of the $n$ gon is $A = \dfrac{1}{4}\cot(\dfrac{\pi}{n})x^2 \implies$

The volume $V$ of the $n$ - gonal pyramid is $V = \dfrac{1}{12}x^2\cot(\dfrac{\pi}{n})H$

Let $h_{1} = \dfrac{1}{2}x_{1}\cot(\dfrac{\pi}{n})$

$\triangle{PST} \sim \triangle{PSR} \implies \dfrac{16}{x_{1}\cot(\dfrac{\pi}{n})} =$ $\dfrac{2H}{x\cot(\dfrac{\pi}{n})}$

$\implies \dfrac{8}{x_{1}} = \dfrac{H}{x} \implies x_{1} = \dfrac{8x}{H}$

$\implies V_{1} = \dfrac{1}{12}x^2\cot(\dfrac{\pi}{n})H - \dfrac{1}{12}(\dfrac{8x}{H})^2\cot(\dfrac{\pi}{n})(8) =$ $\dfrac{1}{12}x^2\cot(\dfrac{\pi}{n})(\dfrac{H^{3} - 512}{H^2})$

Let $h_{2} = \dfrac{1}{2}x_{2}\cot(\dfrac{\pi}{n})$

$\triangle{PUV} \sim \triangle{PRQ} \implies \dfrac{2(H - 2)}{x_{2}\cot(\dfrac{\pi}{n})} =$ $\dfrac{2H}{x\cot(\dfrac{\pi}{n})}$ $\implies \dfrac{H - 2}{x_{2}} = \dfrac{H}{x} \implies$

$x_{2} = \dfrac{(H - 2)x}{H} \implies$ $V_{2} = \dfrac{1}{12}(\dfrac{(H - 2)^2}{H^2})x^2\cot(\dfrac{\pi}{n})(H - 2) =$

$\dfrac{1}{12}\cot(\dfrac{\pi}{n})x^2\dfrac{(H - 2)^3}{H^2}$

$V_{1} = V_{2} \implies H^3 - 512 = H^3 - 6H^2 + 12H - 8 \implies H^2 - 2H - 84 = 0$

Taking the positive root $\implies H = 1 + \sqrt{85} \approx \boxed{10.2195}$