Too long for a comment. 

Let $$A=\left(\begin{array}{rrrrrr}
0 & 1 & 0 & 0 & 0 & 0 \\
1 & 0 & 1 & 0 & 0 & 0 \\
0 & 1 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 & 1 & 0 \\
0 & 0 & 0 & 1 & 0 & 1 \\
0 & 0 & 0 & 0 & 1 & 0
\end{array}\right)$$ be the adjacency matrix of the path graph for $n=6$. The characteristic polynomial of $A^k$ evaluated at $1$ gives the following for $0\leq k\leq 182$ (we take everything over the ring $\mathbb{Z}/6\mathbb{Z}$):
$$[0,1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 5, 1, 3, 4, 1, 1, 5, 1, 1, 1, 2, 1, 5, 1, 5, 3, 1, 4, 1, 1, 1, 1, 5, 1, 4, 1, 5, 1, 3, 1, 1, 4, 5, 1, 1, 1, 5, 1, 2, 1, 5, 3, 1, 1, 1, 4, 1, 1, 5, 1, 1, 1, 2, 1, 3, 1, 1, 1, 5, 4, 1, 1, 5, 1, 5, 1, 2, 3, 1, 1, 1, 1, 1, 4, 5, 1, 1, 1, 5, 1, 0, 1, 1, 1, 5, 1, 1, 4, 5, 1, 5, 1, 5, 3, 4, 1, 1, 1, 1, 1, 5, 4, 1, 1, 5, 1, 3, 1, 4, 1, 5, 1, 1, 1, 5, 4, 5, 1, 5, 3, 1, 1, 4, 1, 1, 1, 5, 1, 1, 4, 5, 1, 3, 1, 1, 1, 2, 1, 1, 1, 5, 1, 5, 4, 5, 3, 1, 1, 1, 1, 4, 1, 5, 1, 1, 1, 5, 4, 3, 1, 1, 1, 5, 1, 4, 1, 5, 1, 5, 1, 5, 0]$$The smallest $k$ where this happens to be $0$ is the 91st entry in the list. Here the matrix is $$
A^{91} = \left(\begin{array}{rrrrrr}
0 & 0 & 0 & 0 & 0 & 5 \\
0 & 0 & 0 & 0 & 5 & 0 \\
0 & 0 & 0 & 5 & 0 & 0 \\
0 & 0 & 5 & 0 & 0 & 0 \\
0 & 5 & 0 & 0 & 0 & 0 \\
5 & 0 & 0 & 0 & 0 & 0
\end{array}\right)
$$
$A^{182}$ is the identity.

Unfortunately, not even the reasoning $\chi(A^k)(1)\neq0$ implies there is no $w$, such that $Aw=w$ seems to apply in the $\mathbb{Z}/6\mathbb{Z}$ situation; for example we have $$A^7\cdot (3,3,3,3,3,3)^t = (3,3,3,3,3,3)^t$$