Polygons with fixed number of axes of symmetry

Question

Let $k$ be a positive integer. Suppose a polygon has exactly $k$ axes of symmetry. How many sides may the polygon have?

A regular $n$-gon has $n$ axes of symmetry, so one answer is $k$. What are other possible answers?

Answer 1

I have the following ideas, which are not checked in details.

I expect that a we can slightly deformate a regular $nk$-polygon in such a way that the obtained polygon will have exactly $k$ axes of symmetry.

From the other side, let $G$ be a group of all symmetries of a polygon $P$ and $G_+$ be the subgroup of the group $G$ consisting of symmetries which do not change the orientation of the polygon $P$. Then $|G:G_+|=2$ and an element $g\in G$ is an axial symmetry iff $g\in G\setminus G_+$. So $|G|=2k$. By Brower theorem, each element $g\in G$ has a fixed point. Thus the elements of the group $G_+$ are rotations. Therefore the action of the group $G_+$ on the set $V$ of vertices of the polygon $P$ has no fixed points. Hence each $G_+$-orbit in the set $V$ consists of $k$ elements and so $|V|$ is divisible by $k$.

PS. I remark that if $k>2$ is then all axis of symmetry of $P$ has a common point and if $k$ is even then $P$ has a center of symmetry.