A problem similar to matrix scaling I'm interested in solutions to the following problem, which is clearly related to the problems known as matrix scaling and matrix balancing (as described in (1), for example), but is different from both of them.
Given a matrix $A$ with non-negative entries, find a positive scalar $Z$ and a diagonal matrix $D$ with positive diagonal entries, such that all the column sums of $\frac{1}{Z} DAD^{-1}$ are equal to unity.
In matrix scaling we're looking for two different diagonal matrices $D_1$ and $D_2$ such that both the row and column sums of $D_1AD_2$ take on specified values; and in matrix balancing we're looking for $D$ such that $DAD^{-1}$ has row sums equal to its own column sums. In this new problem we only want the column sums of $\frac{1}{Z} DAD^{-1}$ to have a specified value rather than being equal to the row sums, and we don't care about the row sums at all.
It obviously isn't possible to solve this for every matrix. In particular, it can't be solved if $A$ is diagonal with unequal diagonal entries, since in this case $\frac{1}{Z} DAD^{-1} = \frac{1}{Z} A$.
So my questions are as follows:


*

*What is the class of matrices for which this problem can be solved?

*When there is a solution, is it typically unique?

*Is there an analytical solution?

*If not, is there an algorithm that can be used to solve it numerically?
Reference:
(1) A Nemirovski, U Rothblum (1999) On complexity of matrix scaling. Linear Algebra and its Applications, 302, 435-460.
 A: Answering my own question again. This one's actually kind of easy.
Let the diagonal elements of $\mathrm{D}$ be written $1/y_i$. Then, writing it out explicitly, the problem is to find numbers $\{y_i\}$ such that
$$
\frac{1}{Zy_i}\sum_j = a_{ij}y_j = 1,
$$
or
$$
\mathrm{A}\mathbf{y} = Z\mathbf{y},
$$
where $\mathbf{y}$ is a vector with elements $y_i$. 
In other words, this was just an eigenvalue problem in disguise. We're looking for an eigenvector $\mathbf{y}$ of $\mathrm{A}$ with positive elements, with $Z$ being its corresponding eigenvalue.
If the elements of $\mathrm{A}$ are strictly positive, the Perron-Frobenius theorem tells us that the solution to this problem always exists (given by the leading eigenvalue and its corresponding eigenvector), that it's unique (up to normalisation of $\mathbf{y}$), and that $Z$ will always be real and positive.
My counterexample had zero elements in $\mathrm{A}$, which leads to zero elements in its leading eigenvector, which prevents it from being used to scale the matrix in such a way.
