First, we multiply both sides by $ \sin \frac{\pi}{n} $: \[\sin \frac{\pi}{n} \cos \frac{\pi}{n} \cos \frac{2 \pi}{n} \cos \frac{4 \pi}{n} \cos \frac{8 \pi}{n} \cos \frac{16 \pi}{n} = \frac{1}{32} \sin \frac{\pi}{n}.\]By the double-angle formula, $ \sin \frac{\pi}{n} \cos \frac{\pi}{n} = \frac{1}{2} \sin \frac{2 \pi}{n}, $ so \[\frac{1}{2} \sin \frac{2 \pi}{n} \cos \frac{2 \pi}{n} \cos \frac{4 \pi}{n} \cos \frac{8 \pi}{n} \cos \frac{16 \pi}{n} = \frac{1}{32} \sin \frac{\pi}{n}.\]We can apply the double-angle formula again, to get \[\frac{1}{4} \sin \frac{4 \pi}{n} \cos \frac{4 \pi}{n} \cos \frac{8 \pi}{n} \cos \frac{16 \pi}{n} = \frac{1}{32} \sin \frac{\pi}{n}.\]Going down the line, we eventually arrive at \[\frac{1}{32} \sin \frac{32 \pi}{n} = \frac{1}{32} \sin \frac{\pi}{n},\]so $ \sin \frac{32 \pi}{n} = \sin \frac{\pi}{n} $. The sine of two angles are equal if and only if either they add up to an odd multiple of $ \pi, $ or they differ by a multiple of $ 2 \pi $. Thus, either \[\frac{33 \pi}{n} = \pi (2k + 1)\]for some integer $ k, $ or \[\frac{31 \pi}{n} = 2 \pi k\]for some integers $ k $. The first condition becomes $ n(2k + 1) = 33, $ so $ n $ must be a divisor of 33. These are 1, 3, 11, and 33. The second condition becomes $ nk = \frac{31}{2}, $ which has no integer solutions. The only step we must account for is when we multiplied both sides by $ \sin \frac{\pi}{n} $. This is zero for $ n = 1, $ and we see that $ n = 1 $ does not satisfy the original equation. Thus, the only solutions are $ \boxed{3, 11, 33} $.