$A\in{\mathbb{R}^+}^{m\times n}$ (i.e. a matrix with elements that are positive real numbers).
The rank of $A$ is $k$, and $k < min(m,n)$.
Therefore $A = BC$, where $B\in{\mathbb{R}^+}^{m\times k}$ and $C\in{\mathbb{R}^+}^{k\times n}$.
Also, we know there exists $D\in{\mathbb{R}^+}^{k\times k}$, with $A=BD^{-1}DC$, and with the elements of $BD^{-1}$, and $DC$ being positive reals (at least one example of D is the identity matrix).
My question is this: what general constraints on the form of $D$ give it the property of being non-singular and the property of keeping $BD^{-1}\in{\mathbb{R}^+}^{m\times k}$ and $DC\in{\mathbb{R}^+}^{k\times n}$.
(Or equivalently I am asking if there is any relationship between different non-negative matrix factorizations).
First, this is not true that any rank-$k$ nonnegative matrix can be written as a product of $m\times k$ and $k\times n$ nonegative matrices. Let's take the matrix $$A=\left(\begin{array}{cccc}1&1&0&0&\\0&1&1&0\\0&0&1&1\\1&0&0&1\end{array}\right)$$ of rank three. Note that any non-negative rank-one matrix $B\leqslant A$ has at most one non-zero diagonal element, so that $A$ cannot be written as a sum of three or less nonnegative rank-one matrices. Or, equivalently, $A$ cannot be written as a product $4\times 3$ and $3\times 4$ nonegative matrices. As to your second question, I think the nonnegative factorization spaces can be very complicated, so I do not know of any result of this kind. But if you wanted $D^{-1}$ to be nonnegative as well, you would need $D$ to be a monomial matrix as it is not hard to see.
Edit. If you do not allow the matrices to contain zeros, you can replace them by small $\varepsilon$'s in $A$, and the same conclusion will hold by almost the same argument.