Since the eigenspace corresponding to $\lambda=0$ is 2-dimensional, there are 2 Jordan blocks for $\lambda=0$; and since this eigenvalue has algebraic multiplicity 4, the two blocks have to have sizes adding to 4.
Therefore there is a $3\times3$ block and a $1\times1$ block, or there are two $2\times2$ blocks.
In the first case, we could find a string $z_i\longrightarrow y_i\longrightarrow x_i$ for one of the eigenvectors, where
the arrow indicates the operation of applying $A-\lambda I$ to the vector;
so since $\lambda=0$, this would mean that $A^2(z_i)=x_i$, and the author claims that this system has no solution (for $i=2$ or $i=3$).
Therefore there must be two $2\times2$ blocks, and so we can find strings
$y_i\longrightarrow x_i$ for $i=2$ and $i=3$ by solving $A y_i=(A-\lambda I)y_i=x_i$ for $y_i$.
The columns of $X$ consist of the vectors in these strings and the eigenvector chosen for $\lambda=1$.
As a reminder of some theory:
Suppose $M$ is an arbitrary $2 \times 2$ matrix with two eigenvalues, $\lambda_1, \lambda_2$, and hence two corresponding eigenvectors, $v_1$ and $v_2$. Define a matrix $V$ whose columns are those two eigenvectors,
\begin{equation} V =\left( \begin{matrix} v_1 & v_2 \\ | & | \end{matrix} \right) \end{equation}
where those vertical bars indicate the column.
Then the action of $M$ on $V$ is the same as the action of $M$ on the individual columns. As those columns are also eigenvectors we may write
\begin{equation} MV =\left( \begin{matrix} Mv_1 & Mv_2 \\ | & | \end{matrix} \right) = \left( \begin{matrix} \lambda_1v_1 & \lambda_2v_2 \\ | & | \end{matrix} \right) = \underbrace{\left( \begin{matrix} v_1 & v_2 \\ | & | \end{matrix} \right)}_{V} \underbrace{\left( \begin{matrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{matrix} \right)}_{D} \end{equation}
where $D$ is the diagonal matrix of eigenvalues. That is, $MV = VD$. Or
\begin{equation} M = VDV^{-1} \end{equation}
This is a change of basis, to/from a basis of eigenvectors.
In this notation, the Jordan normal form is the diagonal matrix $D$. The change of basis matrix $V$ comprises of eigenvectors.
All of this generalizes to $n \times n$ matrices. And there is the added complication that you can have Jordan blocks for eigenspaces where the geometric multiplicity is less than the algebraic multiplicity. However that last wrinkle doesn't apply to your example.
Back to your problem: the basis is an appropriate basis iff it corresponds to the eigenvalues on the diagonal, $\lambda = 5, 3, 3$. To check that, apply the matrix $A$ to each of them and confirm they are eigenvectors with those eigenvalues. (Hint: they are!)
Best Answer
From $A^k=A^t$ we see that $A^tA=A^{k+1}=AA^t$. Hence $A$ is normal, and diagonalizable over $\mathbb{C}$. Hence $J(A)^t=J(A)=J(A^t)=J(A^k)=J(A^k)^t$. $J(A^k)$ and $(J(A))^k$ are only equal if you decide to order the eigenvalues in the corresponding way, and I am not sure there is a canonical way to do that.
In any case, your $A^2=A^t$ question is reduced to the same question for diagonal matrices.