It is easy to find solutions with additional symmetries. Precisely: If $a\neq\cos(\pi m/n)$ there exists exactly one solution such that $|x_k-p_k|=|x_{k+1}-p_k|$ for all $k\in \mathbb{Z} $ (here we denote the sequence $(p_k)$ as an $n$-periodic sequence indicized on $ \mathbb{Z} $).

Indeed, let $u:=e^{i\theta}$ with $\theta=2\arccos a$. By the assumption, $u^n\neq 1$. Starting with $x_0:=x\in\mathbb{C}$, define inductively $x_{k+1}=(x_k-p_k)u+p_k$. The periodicity condition $x_n=x_0$ corresponds then to a non-singular linear equation on $x$ (very easy to write down, which pleasure I won't spoil).

The same procedure produces, more generally, for any $a$ and for any model triangle, with vertices $(0,1 ,u)$ such that $|1-u|=a(1+|u|)$ and $u^n\neq1$, a unique solution such that the triangles with ordered vertices $(p_k,x_k,x_{k+1} )$ are all similar to the fixed triangle with ordered vertices $(0,1,u)$.

It is easy to find solutions with additional symmetries. Precisely: If $a\neq\cos(\pi m/n)$ there exists exactly one solution such that $|x_k-p_k|=|x_{k+1}-p_k|$ for all $k\in \mathbb{Z} $ (here we denote the sequence $(p_k)$ as an $n$-periodic sequence indicized on $ \mathbb{Z} $).

Indeed, let $u:=e^{i\theta}$ with $\theta=2\arccos a$. By the assumption, $u^n\neq 1$. Starting with $x_0:=x\in\mathbb{C}$, define inductively $x_{k+1}=(x_k-p_k)u+p_k$. The periodicity condition $x_n=x_0$ corresponds then to a non-singular linear equation on $x$ (very easy to write down, which pleasure I won't spoil).

The same procedure produces, more generally, for any $a$ and for any model triangle, with vertices $(0,1 ,u)$ such that $|1-u|=a(1+|u|)$ and $u^n\neq1$, a unique solution such that the triangles with ordered vertices $(p_k,x_k,x_{k+1} )$ are all similar to the fixed triangle with ordered vertices $(0,1,u)$.

It is easy to find solutions with additional symmetries. Precisely: If $a\neq\cos(\pi/n)$ there exists exactly one solution such that $|x_k-p_k|=|x_{k+1}-p_k|$ for all $k\in \mathbb{Z} $ (here we denote the sequence $(p_k)$ as an $n$-periodic sequence indicized on $ \mathbb{Z} $).

Indeed, let $u:=e^{i\theta}$ with $\theta=2\arccos a$. By the asumption, $u^n\neq 1$. Starting with $x_0:=x\in\mathbb{C}$, define inductively $x_{k+1}=(x_k-p_k)u+p_k$. The periodicity condition $x_n=x_0$ correspond then to a non-singular linear equation on $x$ (very easy to write down, which pleasure I won't spoil).

The same procedure produces, more generally, for any $a$ and for any model triangle (but one), with vertices $(0,x ,x')$ a unique solution such that the triangles with vertices $(p_k,x_k,x_{k+1} )$ are all similar to the fixed triangle with vertices $(0,x ,x')$.

It is easy to find solutions with additional symmetries. Precisely: If $a\neq\cos(\pi m/n)$ there exists exactly one solution such that $|x_k-p_k|=|x_{k+1}-p_k|$ for all $k\in \mathbb{Z} $ (here we denote the sequence $(p_k)$ as an $n$-periodic sequence indicized on $ \mathbb{Z} $).

Indeed, let $u:=e^{i\theta}$ with $\theta=2\arccos a$. By the assumption, $u^n\neq 1$. Starting with $x_0:=x\in\mathbb{C}$, define inductively $x_{k+1}=(x_k-p_k)u+p_k$. The periodicity condition $x_n=x_0$ corresponds then to a non-singular linear equation on $x$ (very easy to write down, which pleasure I won't spoil).

The same procedure produces, more generally, for any $a$ and for any model triangle, with vertices $(0,1 ,u)$ such that $|1-u|=a(1+|u|)$ and $u^n\neq1$, a unique solution such that the triangles with ordered vertices $(p_k,x_k,x_{k+1} )$ are all similar to the fixed triangle with ordered vertices $(0,1,u)$.