It's easy to see that the set $\{W - W^T : W \in \mathbb{R}^{n \times n}\}$ is precisely the set of real skew-symmetric matrices. This continues to be the case if we restrict to (entry-wise) non-negative matrices (i.e., $$\{W - W^T : W \in \mathbb{R}^{n \times n}\} = \{W - W^T : W \in \mathbb{R}^{n \times n}, \text{ each } W_{i,j} \geq 0\} = \{A \in \mathbb{R}^{n \times n} : A \text{ is skew-symmetric}\}).$$ Is there a simple condition if we restrict to non-negative right-stochastic matrices? (i.e., those whose rows sum to $1$)? That is, is there a simple condition $C$ on $A$ such that $$\{W - W^T : W \in \mathbb{R}^{n \times n}, \text{ each } W_{i,j} \geq 0, W \text{ right-stochastic}\} = \{A \in \mathbb{R}^{n \times n} : A \text{ is skew-symmetric}, C \text{ holds}\})?$$
For context, I reduced a problem to the optimization problem $$\min_{W \in \mathbb{R}^{n \times n}} x^T (W - W^T) y$$ where $x, y \in \mathbb{R}^n$ are fixed vectors, subject to $W$ being non-negative and right stochastic, and I'm wondering whether there's a simple equivalent problem of the form $$\min_{A \in \mathbb{R}^{n \times n} \text{skew-symmetric}} x^T A y.$$
A matrix $K\in\mathbb R^{n\times n}$ is equal to $W-W^T$ for some row-stochastic matrix $W\in\mathbb R^{n\times n}$ if and only if $K$ is a skew-symmetric matrix such that $|k_{ij}|\le1$ for every $(i,j)$ and $\sum_{j=1}^n\max(k_{ij},0)\le1$ for every $i$.
If $K=W-W^T$ for some row-stochastic matrix $W$, then $-1\le-w_{ji}\le w_{ij}-w_{ji}\le w_{ij}\le1$ for every $(i,j)$. Therefore $|k_{ij}|\le1$. Also, $$ \sum_{j=1}^n\max(k_{ij},0)=\sum_{j=1}^n\max(w_{ij}-w_{ji},0)\le\sum_{j=1}^nw_{ij}=1. $$
Conversely, suppose $K$ is a skew-symmetric matrix such that $|k_{ij}|\le1$ for all $(i,j)$ and $\sum_{j=1}^n\max(k_{ij},0)\le1$ for every $i$. Define $W$ by \begin{aligned} w_{ij}&=\max(k_{ij},0)\ \text{ when } i\ne j,\\ w_{ii}&=1-\sum_{j\ne i}\max(k_{ij},0)\ \text{ for each }i. \end{aligned} Then $W$ is row-stochastic and $W-W^T=K$.