I'm not sure if this is an acceptable answer, but it's slick! By the Cayley–Hamilton theorem, $\displaystyle A^4+A+I=0$. Therefore $\displaystyle A = -(A^4+I)$ and so $\displaystyle A^2 = (A^4+I)^2 = A^8+2A^4+I$. Hence $\displaystyle A^8 +2A^4-A^2+I = 0$, which says that $\displaystyle A^2$ satisfies the polynomial equation $\displaystyle t^4+2t^2-t+1=0$. So I assume that $\displaystyle t^4+2t^2-t+1$ is the characteristic polynomial of $\displaystyle A^2$. But I don't know whether that conclusion is justified.