Given a square matrix
$B=\begin{bmatrix}5 & -6 & -6\\-1 & 4 & 2\\3 & -6 & -4\end{bmatrix}$
with eigenvalues $1,2,2$, I am interested to find a rational canonical form of the matrix $B$. In the book of Linear Algebra by Hoffman & Kunje, I found that the rational form of the matrix being
$A=\begin{bmatrix}0 & -2 & 0\\1 & 3 & 0\\0 & 0 & 2\end{bmatrix}=\begin{bmatrix}A_{1} & \mathbf{0}_{2\times{1}}\\\mathbf{0}_{1\times{2}} & 2\end{bmatrix}$,
in which the first block $A_1$ has dimension $2$ corresponding to a cyclic subspace $Z(\alpha_{1},B)$ spanned by the vectors
$\alpha_{1}=\begin{bmatrix}1\\0\\0\end{bmatrix}$ and $B\alpha_{1}=\begin{bmatrix}5 \\-1\\3\end{bmatrix}$.
The last diagonal block corresponds to the eigenspace spanned by
$\alpha_{2}=\begin{bmatrix}2\\1\\0\end{bmatrix}$
for eigenvalue $2$. However, I am curious if I can transform $B$ into a different rational canonical form $B_{R}$ as follows
$B_{R}=\begin{bmatrix}0 & -4 & 0\\1 & 4 & 0\\0 & 0 & 1\end{bmatrix}=\begin{bmatrix}B_{1} & \mathbf{0}_{2\times{1}}\\\mathbf{0}_{1\times{2}} & 1\end{bmatrix}$
where $B_{1}$ is of dimension $2$ corresponding to a cyclic subspace $Z(\gamma_{1},B)$ and the last block corresponding to eigenspace spanned by eigenvector
$\begin{bmatrix}1\\-1/3\\1\end{bmatrix}$ with eigenvalue $1$.
My question is whether such a transformation matrix $T$ exists that $B_{R}={T}^{-1}B{T}?$ If yes, then how to find cyclic vector $\gamma_{1}$. Any hints or suggestions are highly appreciated.