Introduction :
Suppose $n$ is a natural number which is equal to or greater than $5$.
Let's think of regular $n$-gon in any Euclidean space, and denote $x_1, x_2, \ldots, x_n$ the points of regular $n$-gon on the counterclockwise way.
Define $c_n$ by $\frac{sin\frac{3\pi}{n}}{sin\frac{\pi}{n}}$, it is the ratio of second small diagonal (of regular $n$-gon) to one side (of that regular $n$-gon).
Then we can easily get the following equations:
$x_4 - x_1 = c_n (x_3 - x_2)$
$x_5 - x_2 = c_n (x_4 - x_3)$
$\vdots$
$x_3 - x_n = c_n (x_2 - x_1)$
Generally, let's define $a_{i+3}=c_n a_{i+2}-c_n a_{i+1} +a_i$ for any $i \in \mathbb{N}$.
If we assign $a_1 = x_1, a_2 = x_2, a_3 = x_3$, then we get $a_m = x_m$ for all $m \in \lbrace 1,2,\ldots, n \rbrace$, and $a_{n+1}=a_1, a_{n+2}=a_2, a_{n+3}=a_3$.
Question :
If $a_{i+3}=c_n a_{i+2}-c_n a_{i+1} +a_i$ for any $i \in \mathbb{N}$ with any initial values of $a_1, a_2, a_3$, then I guess that the equations $a_{n+1}=a_1, a_{n+2}=a_2, a_{n+3}=a_3$ are also true.
It is equivalent to the following question :
Let $A_n = \pmatrix {0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -c_n & c_n}$, then is $\left(A_n\right)^n$ equal to $I$?
I've checked that it is true where $n=5,6,7$.
It is easy to check when $n=6$. In this case, $c_6 = 2$,
$A_6 = \pmatrix {0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -2 & 2}$,
and $\left(A_6\right)^6 = \pmatrix {0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & -2 & 2}^6 = I$.
Motivation :
If this conjecture is true, then the following interesting result is also true.
In any Euclidean space, choose any three points $P_1, P_2, P_3$, and choose any $c_n$ which is defined in Introduction. Then for $i=4,5,6,\ldots$, define $P_i$ recursively by $P_i - P_{i-3} = c_n (P_{i-1}-P_{i-2})$.
It means that $\overline{P_i P_{i-3}}$ is parallel to $\overline{P_{i-1} P_{i-2}}$, and the ratio of $\overline{P_i P_{i-3}}$ to $\overline{P_{i-1} P_{i-2}}$ is $c_n$ for any $i = 4,5,6, \ldots$.
Then surprisingly $P_{n+1}$ is $P_1$, $P_{n+2}$ is $P_2$, and $P_{n+3}$ is $P_3$.
Note that $c_n=2\cos(2\pi/n)+1$ and that $A_n$ is the companion matrix of the polynomial $x^3-c_nx^2+c_nx-1=(x-1)(x^2-(c_n-1)x+1)=(x-1)(x^2-2\cos(2\pi/n)x+1)$.
To show that $A_n^n=I_3$, it suffices to show that the roots of $x^2-2\cos(2\pi/n)x+1$ are not $1$, are distinct and satisfy $x^n=1$. That is the case because the considered roots are $e^{\pm 2i\pi/n}$.