= Solution
The inverse appearing in the assumption requires $\det C\ne0$. Taking <determinants> of $(BC)^{\mathsf T}=-C^{-1}B$ gives
$$
\det B\,\det C=-\frac{\det B}{\det C},
$$
where the minus sign comes from the odd <matrix> order. Multiplying by $\det C$ gives $\det B((\det C)^2+1)=0$. Since $C$ is real, the second factor is positive. Thus $\boxed{\det B=0}$ and $B$ is a <singular matrix>. This is the <singularity forced by an odd-dimensional transpose-inverse identity>, with the roles of the two <matrix> letters exchanged.
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