I have such an equation $(X^{-1} \cdot 2 \cdot A)^{-1} - B = 4X $ Where $B$ and $A$ are matrices.
So, I did such a simplification.
\begin{align} & \frac 12 A^{-1}X - B = 4X\\ \iff & \frac 12 A^{-1} X = 4X + B\\ \iff & \frac 12 A^{-1} X - 4X = B\\ \iff & \left(\frac 12 A^{-1} - 4I\right)X = B\\ \iff & X = B \left(\frac 12 A^{-1} - 4I\right)^{-1}. \end{align}
Am I right?
Not really.
It is correct that $(X^{-1}\cdot2\cdot A)^{-1}=\frac{1}{2}A^{-1}X$, so the equation becomes $$ \frac{1}{2}A^{-1}X-4X=B $$ It's easier if you simplify it to $A^{-1}X-8IX=2B$, so $$ (A^{-1}-8I)X=2B $$ and therefore, multiplying by the inverse on the left, $$ X=2(A^{-1}-8I)^{-1}B $$ Your solution would be good as well, provided you keep the correct side of the multiplication.
In order to do those steps you need that $A$ is invertible and hasn't $1/8$ among its eigenvalues.