= Solution
The cyclic <matrix> acts by $(v_1,v_2,v_3,v_4)\mapsto(v_2,v_3,v_4,v_1)$. For each fourth root of unity $\lambda$, the <vector> $v_\lambda=(1,\lambda,\lambda^2,\lambda^3)^T$ satisfies $Av_\lambda=\lambda v_\lambda$, because $\lambda^4=1$. This supplies the four distinct <eigenvalues> and nonzero <eigenvectors>:
$$
\begin{array}{c|c}
\lambda&v_\lambda^T\\\hline
1&(1,1,1,1)\\
-1&(1,-1,1,-1)\\
i&(1,i,-1,-i)\\
-i&(1,-i,-1,i)
\end{array}
$$
<Eigenvectors> for distinct <eigenvalues> are linearly independent. Thus, with these columns in the indicated order,
$$
P=\begin{pmatrix}1&1&1&1\\1&-1&i&-i\\1&1&-1&-1\\1&-1&-i&i\end{pmatrix},\qquad
\boxed{P^{-1}AP=\operatorname{diag}(1,-1,i,-i).}
$$
They form a <basis> over $\mathbb C$, so there are no other <eigenvalues>. Equivalently the <characteristic polynomial> is $\lambda^4-1$.
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