= Solution
For either $A$ from part (i), choose
$$
\boxed{B=-\frac{I+A}{\sqrt2}.}
$$
Indeed, using $A^2=-I$ and $A^{\mathsf T}=-A$,
$$
B^2=\frac{I+2A+A^2}{2}=A,\qquad
B^{\mathsf T}B=\frac{(I-A)(I+A)}2=I.
$$
Its <matrix trace> is $-2/\sqrt2=-\sqrt2<0$. Explicitly, for the two $A$ matrices in the order displayed above, the corresponding <orthogonal matrices> are
$$
\boxed{B=\frac1{\sqrt2}\begin{pmatrix}-1&1\\-1&-1\end{pmatrix},\qquad
B=\frac1{\sqrt2}\begin{pmatrix}-1&-1\\1&-1\end{pmatrix}.}
$$
Their negatives have positive <matrix trace> and would not meet the additional sign condition.
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