I am trying to solve the matrix equation of the form $XA = XB$. $A$, $B$ and the solution sought $X$ are $4 \times 4$ homegeneous matrices which are composed of a rotation matrix and translation vector, like in $ \begin{bmatrix} r_1 & r_2 & r_3 & tx\\ r_4 & r_5 & r_6 &ty\\ r_7 & r_8 & r_9 & tz\\ 0 & 0 & 0 & 1 \end{bmatrix}.$ There are several methods to solve equations of the type AX = XB. These type of equations occur in domains like robotics, for which solutions have been proposed like the Tsai and Lenz method [IEEE 1987], for example. 1) I feel solving the equation XA = XB is not the same as solving the known form AX = XB. Am I right? 2) Neither can it be solved like the Sylvester equation, because even that requires $AX + XB = C$ form. What i have is $XA = XB$, a redundant set of equations. That is,
\begin{align} A_1.X & = X.B_1 \\ A_2.X & = X.B_2 \\ & \vdots \\ A_n.X & = X.B_n \end{align}
If i am correct, these could be rewritten into another form like $AX = BX$. Should i rewrite the equation and try to solve this problem using any other existing methods?