For $x \in \mathbb{C}$ define $A,B \in M(3\times3, \mathbb{C})$ as $$ A = \begin{pmatrix} x & 0 & 0 \\ 0 & x & 1 \\ 0 & 0 & x \end{pmatrix}$$ and $$ B = \begin{pmatrix} x & 1 & 0 \\ 0 & x & 0 \\ 0 & 0 & x \end{pmatrix}. $$ Now I want to find a transformation matrix $T \in \text{GL}_3(\mathbb{C})$ with $B = T^{-1}AT$.
I am able to perform row/column manipulations to get from A to B, but I yield a wrong result when trying to combine those manipulations via elementary matrices. What is the right way to come up with the matrix T?
Product from the right = operation on columns, product from the left = operation on rows, so take
$$T=\begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&0&1\\0&1&0\end{pmatrix}=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}\implies T^{-1}=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}\;\;,\;\;\;\text{and}$$
$$T^{-1}AT=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}\begin{pmatrix}x&0&0\\0&x&1\\0&0&x\end{pmatrix}\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix}=$$$${}$$
$$=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}\begin{pmatrix}0&0&x\\x&1&0\\0&x&0\end{pmatrix}=\begin{pmatrix}x&1&0\\0&x&0\\0&0&x\end{pmatrix}=B$$