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The problem is as follows:

Find at least one polygon $\mathbb{P}$ whose group of symmetries has order $n \in \mathbb{N}$ and is equal to its group of rotations. Is that group cyclic?

Although the problem doesn't specify how many vertices the polygon should have, I assume that it should be an $n$-gon. What confuses me is that in order for the polygon to have $n$ rotations, it needs to be cyclic, and all of its central angles must be equal, which means that it must be regular, but regular polygons have $2n$ symmetries. Is my conclusion wrong? If so, which part of my deduction is wrong? I'd appreciate any input, and of course, a solution (even without any attention to my additional questions, which might be completely off the mark).

EDIT: In case it's unclear, the group of symmetries of a polygon $\mathbb{P}$ is the set of all isometries $\phi:\mathbb{E}^2 \to \mathbb{E}^2$ such that $\phi(\mathbb{P})=\mathbb{P}$ with its operation being function composition. Also, its group of rotations (almost always) consists of rotations around the same point, as the composition of two rotations around different points need not be a rotation.

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    Taking the obvious risk... imagine a swastika with $n$ hooks instead of the usual four. And, no, the assumption that the polygon should be an $n$-gon is the one that lead you astray. – 2017-02-25
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    So what would the polygon be, for example, for $n=3$? I'm imagining a triangle whose vertices are the ends of a 3-hook swastika-like figure. Or is it maybe the hexagon which includes the vertices of the triangle, but also has another 3 vertices, the points where the hooks "break"? – 2017-02-25
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    A famous example is the [triskelion](https://en.wikipedia.org/wiki/Triskelion), although it's not polygonal (interestingly, I forgot its name, but remembered it was on the flag for the Isle of Man -- weird!). It shouldn't be too hard to imagine a polygonal version; start with an equilateral triangle, and attach smaller equilateral triangles to each edge, near the vertices of the original triangle (I have a $9$-gon version in mind, for $n = 3$ symmetries. And more generally, a $3n$-gon, for $n$ symmetries). – 2017-02-25

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