Solve a matrix equation of the form $A=XB$ 
Given matrices $A \in \mathbb{R}^{n \times p}$ and $B \in \mathbb{R}^{m \times p}$, is it possible to solve the matrix equation $A=XB$?

I don't know if this is feasible. It may even be quite easy in fact but I'm stuck with that equation in my work (I'm not a math specialist). I've tried to make the inverse of $B$, but as $B$ is not a square matrix in my data, I can't do that. In my data $A$ is $120 \times 170$ and $B$ is $10 \times 170$. I am expecting $X$ to be $120 \times10$. Any ideas?
EDIT:  matrix $B B^\top$ is invertible. Giving a $10 \times 10$ matrix.
 A: There can not be guaranteed solution. This is evident by the fact, that $A$ can has a rank up to $p$, while the rank of $XB$ is limited by $p$ (if we keep the order of your sizes). 
If you draw a picture of your matrices (just the boxes) you find, that you are searching for a low rank representation of $A$ with a specific set of right singular vectors. I italize because these are most likely not orthogonal. 
You can approximate $A$ by your product. The best (I assume) way to do this, is by choosing the pseudo-inverse of $B$. (Matlab gives this by pinv). 
A: The system has solution $\iff Row(A)\subseteq Row(B)$ thus we can find a basis for $Row (B)$ and check that the condition is matched.
If $Row(A)\subseteq Row(B)$ we can obtain $X$ solving the system row by row
$$R_i(A)=R_i(X)B$$
A: The way to think of this is as a "multiple right-hand side problem"
$$A = X B $$
can be thought of as 
$$
\left( \begin{array}{c}
a_0^T \\
a_1^T \\
\vdots
\end{array}
\right)
=
\left( \begin{array}{c}
x_0^T \\
x_1^T \\
\vdots
\end{array}
\right) B
=
\left( \begin{array}{c}
x_0^T B\\
x_1^T B\\
\vdots
\end{array}
\right)
$$
where $ a_i^T $ and $ x_i^T $ are the rows of $ A $ and $ X $ indexed with $ i $, respectively.
What you recognize is that you are solving many $ a_i^T = x_i^T B $ problems.
Transpose both sides and you get $ B^T x_i = a_i $, which should look a lot more familiar, and then you can extend everything you know about solving such a system.
