Let $n \in \mathbb{N}$. I am looking for an explicit example of an open, bounded, symmetric, convex subset $C \subset \mathbb{R}^2$ whose full Euclidean symmetry group (that is, the group of isometries of $\mathbb{R}^2$ mapping $C$ onto itself) is exactly the cyclic group $\mathbb{Z}_n$. In particular, I want no reflection symmetries (i.e. not the full dihedral group $D_n$).
Does there exist a simple, rigorous construction of such a set?
Consider the cyclic polygon whose vertices are $(\cos \theta, \sin \theta)$ where $\theta$ is of the form $\frac{2\pi}{n} (k)$, $\frac{2\pi}{n} (k + \frac{1}{2})$, or $\frac{2\pi}{n} (k + \frac{5}{6})$ where $0 \le k < n$. Then the interior of that polygon will satisfy the required conditions.
The idea here is that the sides subtend central angles of $\frac{2\pi}{n} (\frac{1}{2})$, $\frac{2\pi}{n}(\frac{1}{3})$, $\frac{2\pi}{n}(\frac{1}{6})$, $\frac{2\pi}{n} (\frac{1}{2})$, $\ldots$. Any symmetry of the interior of the polygon must also form a symmetry of the polygon itself. That symmetry must take the first side with central angle of $\frac{2\pi}{n}(\frac{1}{2})$ to one of the $n$ sides with that central angle. An orientation-reversing symmetry is forbidden since the second side, with central angle $\frac{2\pi}{n}(\frac{1}{3})$, must be mapped to the following side, and not the preceding side which has central angle $\frac{2\pi}{n}(\frac{1}{6})$ instead.
