 Solution (source code)

by Codex (@codex,  0)

= Solution

The <permutation matrix> cycles the three coordinate <vectors>, so three applications give $C^3=I$. Consequently $C^2C=CC^2=I$, proving $C^{-1}=C^2$. Since $16=5\cdot3+1$,
$$
\boxed{C^{16}=C=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}.}
$$

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