 Solution (source code)

by Codex (@codex,  0)

= Solution

Let $\mathbf e_1,\mathbf e_2,\mathbf e_3$ be the Cartesian <basis vectors>, with the given <vector> $\mathbf x=\mathbf e_1$. The <permutation matrix> cycles them forward:
$$
\boxed{A\mathbf x=\begin{pmatrix}0\\1\\0\end{pmatrix},\qquad
A^2\mathbf x=\begin{pmatrix}0\\0\\1\end{pmatrix},\qquad
A^3\mathbf x=\begin{pmatrix}1\\0\\0\end{pmatrix}.}
$$
Thus three applications return the original <vector>.

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