If $v_1,\ldots,v_n$ are independent eigenvectors with $Av_j=\lambda_jv_j$, let $P=[v_1\ \cdots\ v_n]$. Then $AP=P\operatorname{diag}(\lambda_1,\ldots,\lambda_n)$, so $P^{-1}AP$ is that diagonal matrix. Reordering columns requires the same reordering of diagonal entries.
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