orthogonal real matrix means characteristic polynomial satsifies $p(t) = t^n p(t^{-1})$

Solution 1:

Hint: Let $f(t)=|t-s|$ with $|s|=1$. Then $tf(\frac 1 1t)=|1-st| =|\frac 1 s -t|=|\overline s -t|=f(t)$ for $t >0$. If $s$ and $\overline s$ are both eigen values then $$(t-s)(t-\overline s)=|t-s|^{2}$$ $$=f(t)^{2}=t^{2}f(\frac 1 t)^{2}$$ $$=t^{2}(\frac 1 t-s) \frac 1 t-\overline s).$$ Apply this to each pair of eigen values and multiply.

This proves that $p(t)=t^{n}p(\frac 1t)$ for $t>0$. Now the identity theorem of Complex Analysis shows that this equation holds for all non-zero complex numbers $t$.

Note that this argument works for conjugate pairs of eigen values. But if $1$ is an eigen value then the argument fails for $f(t)=(t-1)$ since $tf(\frac 1t)=-f(t)$ in this case. So an appropriate adjustment in the sign is needed at the end.

Solution 2:

You need to add an additional condition $1=(-1)^n {\rm Det}(Q)$.

With this condition, you have almost proven it already: As the roots of $p(t)$ (other than $1,-1$) come in conjugate pairs and lie in the unit circle, they also come in inverse pairs, as the conjugate of a unit complex number is its inverse: $(x+iy)(x-iy)=x^2+y^2$. Thus if $p(\lambda)=0$ then $p(\lambda^{-1})=0$.

Thus the polynomials $p(t)$ and $t^np(t^{-1})$ have the same roots - counted with multiplicity. It remains to check they have the same leading coefficient. However this follows from our additional condition: $1=(-1)^n {\rm Det}(Q)$.

Without the additional condition we have a $1\times 1$ counterexample: $\left(1\right)$

That is: $t-1\neq t(t^{-1}-1)$.