Let $A \in \mathbb{R}^{m \times m}$. Give necessary and/or sufficient conditions for the existence of a matrix $D \in \mathbb{R}^{m \times m}$ such that all eigenvalues of $AD$ have negative real part (i.e., $AD$ Hurwitz). Some initial thoughts ...
Obviously $A, D$ must have full rank.
If $-A$ is so-called $D$-stable, the result seems to follow, although this is a much stronger condition than I need as it ensures that $AD$ is Hurwitz for all positive diagonal matrices $D$ (I think theres an easy extension there to allow for $D$ to have both positive and negative diagonal elements).
An equivalent formulation is that this is true if and only if there exists a $D$ and a matrix $P \succ 0$ such that $P(AD)^{\sf T} + (AD)P \prec 0$. Letting $K = DP$ this is equivalent to the convex inequality $K^{\sf T}A + AK \prec 0$ in the variable $K$. If this is feasible for some $K^{\star}$, you can then probably argue about the factorization $K^{\star} = DP$ with $P \succ 0$ and recover an appropriate choice of $D$?
Note: This matrix analysis question is motivated by a problem in systems and control theory.
Your problem is equivalent to the state feedback problem for LTI systems, so
$$ \dot{x} = \mathcal{A}\,x + \mathcal{B}\,u $$
with $u = -K\,x$ such that $\mathcal{A} - \mathcal{B}\,K$ is Hurwitz. In your case $\mathcal{A}=0$ and $\mathcal{B} = -A$. When $\mathcal{A}=0$ then the pair $(\mathcal{A},\mathcal{B})$ is stabilizable if and only if $\mathcal{B}$ is full rank. The same conditions holds for your problem, so $A$ has to be full rank.