With matrix A, how to calculate matrix B such that BAB-1 equals a specific form? With a matrix $$A =  \left[
\begin{array}{}
  1&2&1&2\\
  0&1&3&4\\ 
0&0&3&5\\
0&0&0&4
\end{array}
\right] $$
I am required to find a matrix B such that $$BAB^{-1 }= \left[
\begin{array}{}
  1&2&0&0\\
0&1&0&0\\
0&0&3&5\\
0&0&0&4
\end{array}
\right] $$
I am not sure on how to begin solving this, do I let B be an unknown matrix and sub it into the equation to solve for it? However, this method seems a bit too long and unwise so any help would be appreciated! Thank you!
 A: As Dietrich points out, this amounts to finding an invertible solution $B$ to $AB = BC$, where
$$
C = \pmatrix{1&2&0&0\\
0&1&0&0\\
0&0&3&5\\
0&0&0&4}.
$$
We can make Dietrich's suggested approach a bit more efficient if we "guess" that there exists a $B$ of the form
$$
B = \pmatrix{I & X\\ 0 & I},
$$
where each block in this partition is of size $2 \times 2$. Notably, $B$ is automatically invertible for any choice of $X$. Denote
$$
A_{11} = \pmatrix{1 & 2\\ 0 & 1}, \quad A_{12} = \pmatrix{1&2\\3&4}, \quad A_{22} = \pmatrix{3&5\\0&4}.
$$
With block-matrix multiplication, the equation $AB = BC$ becomes
$$
\pmatrix{A_{11} & A_{11}X + A_{12}\\0 & A_{22}} = \pmatrix{A_{11} & X A_{22}\\0 & A_{22}} \implies\\ A_{11}X - XA_{22} = -A_{12}.
$$
From there, we could write out all the equations and solve the resulting system on $4$ variables. We can do this efficiently by using the relationship between vectorization and the Kronecker product to quickly arrive at the formulation
$$
(I \otimes A_{11})\operatorname{vec}(X) - (A_{22}^T \otimes I)\operatorname{vec}(X) = \operatorname{vec}(-A_{12}) \implies\\
\left[\pmatrix{A_{11} & 0\\0 & A_{11}} - \pmatrix{3 I & 0\\5 I & 4 I}\right] \pmatrix{x_{11}\\x_{21}\\x_{12}\\ x_{22}} = \pmatrix{-1\\-3\\-2\\-4}.
$$
You should arrive at the solution
$$
X = \frac 1{18}\pmatrix{36&-62\\27&-21}.
$$
