= Solution
The expression assumes $Y$ is invertible. By the <cyclic property of the trace>,
$$
\operatorname{Tr}(YXY^{-1})=\operatorname{Tr}(XY^{-1}Y)=\operatorname{Tr}X.
$$
For a real or complex <antisymmetric matrix>, $X^T=-X$ forces every diagonal entry to satisfy $x_{ii}=-x_{ii}$, hence $x_{ii}=0$. Therefore
$$
\boxed{\operatorname{Tr}Z=\operatorname{Tr}X=0.}
$$
This also demonstrates invariance of the <matrix trace> under <matrix similarity>. Antisymmetry of $Z$ itself is not needed.
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