Let $B,C$ be arbitrary matrices of dimensions $n_2\times n_3$ and $n_1\times n_3$, respectively. What are the conditions for the existence of some $A$ of dimensions $n_1\times n_2$ such that $AB=C$? And if we can prove the existence of such $A$, can we write an explicit expression for it?
In the case $n_3=1$ this amounts to asking, given vectors $x\in\mathbb R^{n_2}$ and $y\in\mathbb R^{n_1}$, whether there is $A$ such that $Ax=y$. This is simple to answer: any $A$ such that $x\in\mathrm{supp}(A)$ and $y\in\mathrm{range}(A)$ will do. One such matrix is $A=yx^*$ (here, $x^*$ denotes the dual of $x$ and $yx^*$ is the outer product). I'm still not sure if there is a good way to write a general form for the full set of solutions for $A$.
What about the general case? Clearly we must have $\mathrm{ker}(B)\subseteq\mathrm{ker}(C)$ and thus $\mathrm{supp}(C)\subseteq\mathrm{supp}(B)$, and also $\mathrm{range}(A)=\mathrm{range}(C)$. This implies that $\min(n_1,n_3)\le \min(n_2,n_3)$, but this is not enough to answer the question.
I don't think whether the matrices are real or complex matters here, but if it does, I would be interested in the solution in both cases.
Note that for the case that $A$ is a single row we have $$ (a_{11}\ a_{12}\ \dots\ a_{1,n_2}) \begin{pmatrix} —\ B_1\ —\\ —\ B_2\ —\\ \vdots \\ —\ B_{n_2}\ — \end{pmatrix} = a_{11}B_1+a_{12}B_2+\cdots+a_{1,n_2}B_{n_2}, $$ where $B_i$ denotes the $i$th row of $B$. Hence, for $n_1=1$ the product $AB$ is a linear combination of the rows of $B$ and $AB=C$ has a solution (in $A$) if and only if the row $C$ is a linear combination of the rows of $B$.
For $n_1>1$, the equation $AB=C$ splits up into $n_1$ equations, one for each row.
Therefore, in general, the equation $AB=C$ has a solution in $A$ if and only if each row of $C$ is a linear combination of the rows of $B$. This might be expressed as $$ \operatorname{rowspace}(C) \subseteq \operatorname{rowspace}(B), $$ or by taking orthogonal components as $$ \ker(C) \supseteq \ker(B). $$ Or as a characterization using ranks: $$ \operatorname{rank}(B) = \operatorname{rank}\begin{pmatrix} B\\C\end{pmatrix}. $$