If $\det(A-\lambda I)=\det(A^{-1}-\lambda I)$, then characteristic polynomial is coefficient symmetry

$det(A-\lambda I)=det(A^{-1}-\lambda I)$ suppose $\lambda=0$, you deduce that $det(A)=det(A^{-1})=1$.

Write $P(\lambda)=det(A-\lambda I)$

$det(A-\lambda I)=det(\lambda A)det({1\over\lambda}I-A^{-1})=\lambda^{2n}P({1\over \lambda})$

Comparing the coefficients of $\lambda^{2n}P({1\over \lambda})$ and $P(\lambda)$ you obtain the result.