Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 1 / b / i

A symmetric matrix that is also an orthogonal matrix satisfies $C=C^{-1}$. Multiplying by the commuting invertible factors gives $(I-B{)}^2=(I+B{)}^2$, hence $4B=0$. Thus there is no nonzero choice of $B$. The only symmetric case is $B=0$, giving $C=I$.