In the real field. Given a diagonal matrix $D$ and a symmetric matrix $A$. For every skew symmetric matrix $S$, is there always a symmetric matrix $H$ such that $-\operatorname{trace}(DSAS)=\operatorname{trace}(DHAH)$? 


If $A$ is also diagonal, this can be easily seen true. 



Existence of a  symmetric matrix.
Maybe I'm missing something, but let $f(H)=\mathrm{tr}(DHAH)$, then  

$$f(\alpha H)=\alpha^2f(H)$$  

for every real $\alpha$. Obviously $f(0)=0$. So the problem boils down to find an $H$ such that $f(H)>0$ and another one such that $f(H)<0$. 


Now, denoting by $h\_k$ the $k-$th column of $H$,  

$$(HAH)\_{ij}=(h\_i)^tAh\_j$$  

Therefore $f(H)=\sum d\_i (h\_i)^tAh\_i$, if $D=\mathrm{diag}(d\_1,\ldots, d\_n)$.


If, wlog, $d\_1>0$ and $d\_2<0$, then we are obviously done.  

If every $d\_i$ has the same sign, but $A$ has a positive and a negative eigenvalue, then again we are done. 


If every $d\_i$ is wlog positive (or null) and $A$ is semi-definite, then $f(H)\geq0$ for every $H$ symmetric and $f(S)\leq 0$ for every $S$ skew-symmetric, so, given $H$ such that $f(H)\neq0$, we can always find $\alpha$ such that $f(\alpha H)=-f(S)$.


