orthogonal matrix with negative determinant

Question

Question: Assume that $A$ is a $n\times n$ orthogonal matrix such that $det(A)<0$. Prove that there exists a nonzero vector $x\in\mathbb{R}^n$ such that $Ax=-x$.

Since $A$ is orthogonal and $det(A)<0$, its determinant is $-1$.

Please help, thanks a lot!

Answer 1

$$\det(A+I)=\det(A)\det(A^{-1}+I)=-\det(A^\top+I)=-\det(A+I)$$