<s>[INST] What's the Jordan form of $J^2$? $$J^2=\begin{pmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{pmatrix}$$
What's the Jordan form of $J^2$?
I know that it has two blocks and $2$ independent eigenvectors.  
So it could be 
$$\begin{pmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}
\text{ or } 
\begin{pmatrix}0&1&0&0&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{pmatrix}$$
But why can't the Jordan chain have the length of $4$ and $1$?
 [/INST] My first answer would have been any of the three existing answers. But here is another take. The Jordan form is unique up to permutation of the blocks. So if we achieve a Jordan form via similarities, it will the the Jordan form. 
One particular case of similarity is to conjugate with permutation matrices. The net effect of conjugating with a transposition is to exchange both the row and the column. So 
$$
\begin{bmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\\ 0&0&0&0&0\end{bmatrix}
\xrightarrow{3\leftrightarrow4}\begin{bmatrix}0&0&0&1&0\\ 0&0&1&0&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix}
\xrightarrow{1\leftrightarrow2}\begin{bmatrix}0&0&1&0&0\\ 0&0&0&1&0\\ 0&0&0&0&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix}
\xrightarrow{2\leftrightarrow3}\begin{bmatrix}0&1&0&0&0\\ 0&0&0&0&0\\ 0&0&0&1&0\\ 0&0&0&0&1\\ 0&0&0&0&0\end{bmatrix}.
$$
At this stage, we have the Jordan form. 
</s>