**UPDATE** Sorry, previous version was wrong. For $n=m=2$, this computation shows that there is such a matrix. **FURHTER UPDATE** However, for $m=n=3$, it shows there isn't.

Taking $n=m=2$, we see that $X$ and $Y$ are of the form $\begin{pmatrix} x & 1-x \\ 1-x & x \end{pmatrix}$ and $\begin{pmatrix} y & 1-y \\ 1-y & y \end{pmatrix}$. Set $W = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ Then $X^T W Y =: \begin{pmatrix} e & f \\ g & h \end{pmatrix}$ with

$$f=c+(a-c)x+(d-c) y+(-a+b+c-d)xy$$
$$g=b+(a-b)x+(d-b) y+(-a+b+c-d)xy$$

Your desire is that we have $f(x,y) = g(x,y) =0$ iff $x=y$. Plugging in $x=y$, we deduce that we must have $b=c=a+d=0$. ~~But then $f$ and $g$ vanish for all $x$ and $y$~~. And, indeed, $(a,b,c,d) = (1,0,0,-1)$ solves the problem for $m=n=2$.

Now run the same analysis with $m=n=3$, thinking about $X$ and $Y$ of the form $\begin{pmatrix} x & 1-x & 0 \\ 1-x & x & 0 \\ 0 & 0 & 1 \end{pmatrix}$. We deduce that $w_{12} = w_{21}=0$ and $w_{11}=-w_{22}$. Similarly, $w_{22} = - w_{33}$, $w_{11} = - w_{33}$ and $w_{13}=w_{31} = w_{23} = w_{32}=0$. But the only solution to these linear equations is $W=0$..