 Solution

The inverse appearing in the assumption requires $\det C\ne 0$. Taking determinants of $(BC{)}^{\mathsf{T}}=-C^{-1}B$ giveswhere 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.