Regular polygon made from the midpoints of the sides of a polygon For what natural numbers n is it true that whenever the midpoints of the sides of a convex n-sided polygon $K_n$ form a regular n-gon, then $K_n$ itself is also a regular n-gon?
I know it is true for $n=3$, and that it isn't true for $n=4$, but I'm stuck with $n>4$.
 A: Let us work in the plane identified with the set of complex numbers $\mathbb{C}$.
Let us prove it is false for an even number of points.
Let us call $v_k \in \mathbb{C} (k=1\cdots n)$ the vertices of the polygon and $m_k \in \mathbb{C}$ the midpoint of $[v_k,v_{k+1}] $ with $v_{n+1}=v_1$. The $m_k$ are known, on a regular polygon, and the $v_k$ are unknown.
Let us call $S_k$ the (central) symmetry wrt $m_k$. Its expression is $Z=S_k(z)=2m_k-z$
(explanation: $Z=2m_k-z \Leftrightarrow \dfrac{z+Z}{2}=m_k$)
If, for example, $S_1$ is followed by $S_2$, one has:
$S_2(S_1(z))=2m_2-(2m_1-z)=2(m_2-m_1)+z$
More generally, by composing  $S_1$ followed by $S_2 \cdots S_n$, (a transformation that we will call $S$) when $n$ is even, one has, by an immediate recurrence:
$$S(z)=2(\sum(-1)^k m_k)+z=z \ \ (1) \ \ \text{which boils down to} \ \ S(z)=z$$
(last equality because the sum is zero for a regular polygon). 
Thus, whatever the initial $z=v_1$, one will be able to construct a closed polygon. It is clear that if one takes $z_1$ very close from $m_1$, $v_2=S_1(v_1)$ will be as well close to $v_1$ but far away from $v_3$, thus generating a non-regular polygon (see image below).

Remark: there is a funny 3D interpretation of the above figure as a perspective view of a particular section of a cube by a plane passing by certain midpoints.
