Assign to the $n$ nodes of a path graph vertex weights forming a permutation of $(0,\ldots,n{-}1)$. Now iterate the following update repeatedly: Each node sums the weights of its neighbors, and that node's weight is replaced (in the next iteration) by the sum $\bmod n$. Here is the start of an example for $n=4$:
[![Path4][1]][1]
The leftmost node just copies the $1$ of its neighbor.
The 2nd node from the left is replaced with $(3+0) \bmod 4$.
Continuing, we fall into a cycle of length $6$: $$ \left( \begin{array}{cccc} 3 & 1 & 0 & 2 \\ 1 & 3 & 3 & 0 \\ 3 & 0 & 3 & 3 \\ 0 & 2 & 3 & 3 \\ 2 & 3 & 1 & 3 \\ 3 & 3 & 2 & 1 \\ 3 & 1 & 0 & 2 \\ \end{array} \right) $$ With a different starting permutation, the cycle length can be $3$: $$ \left( \begin{array}{cccc} 2 & 3 & 1 & 0 \\ 3 & 3 & 3 & 1 \\ 3 & 2 & 0 & 3 \\ 2 & 3 & 1 & 0 \\ \end{array} \right) $$ Similarly, for $n=5$, cycles of length $2$ and $8$ occur. But I was surprised to find that for $n=6$, it appears the process always results in a cycle of length $182$.
Q. What explains cycles of length $182$ for paths of $6$ nodes?
More generally, what explains the cycle lengths for different $n$?
Here are the cycle lengths I've explored so far, verified by
exhaustive search up to $n=9$. The $n=9$ cycle length of $48$ is prominent
among random permutations but I am not certain there is no other cycle length.
See Moritz Firsching's comments and postings for data beyond my calculations.
$$
\left(
\begin{array}{cccccccc}
2 & 3 & 4 & 5 & 6 & 7 & 8 & 9
\\
2 & \{1,4\} & \{3,6\} &
\{2,8\} & 182 & \{6,12\} &
28 & \{24,48\} \\
\end{array}
\right)
$$
I've also explored cycle graphs instead of path graphs.
For a cycle graph of $n=6$ nodes, the process falls into cycles of
length $2$ or $6$. For $n=8$, the cycle length is $1$—all zeros.
[![Cyc6][2]][2]
Iteration falls into a cycle of length $2$.
The process may be similarly defined on any graph. I've made some preliminary explorations without seeing a clear pattern.
Replying to Moritz Firsching's question in a comment, a cycle of length $6$ for $n=7$: $$ \left( \begin{array}{ccccccc} 4 & 5 & 6 & 0 & 1 & 2 & 3 \\ 5 & 3 & 5 & 0 & 2 & 4 & 2 \\ 3 & 3 & 3 & 0 & 4 & 4 & 4 \\ 3 & 6 & 3 & 0 & 4 & 1 & 4 \\ 6 & 6 & 6 & 0 & 1 & 1 & 1 \\ 6 & 5 & 6 & 0 & 1 & 2 & 1 \\ 5 & 5 & 5 & 0 & 2 & 2 & 2 \\ 5 & 3 & 5 & 0 & 2 & 4 & 2 \\ \end{array} \right) $$