Prove that A is zero matrix

Note that the eigenvalues of $A^k + I$ are of the form $\lambda^k + 1$ for all eigenvalues $\lambda$ of $A$. On the other hand, the eigenvalues of a unitary matrix must have absolute value $1$. Thus, each eigenvalue $\lambda$ of $A$ satisfies $|\lambda^k + 1| = 1$ for $k = 1,2,3$.

I claim that the only $\lambda$ for which this holds is $\lambda = 0$. Thus, $A$ has $0$ as its only eigenvalue.

On the other hand, because $A + I$ is unitary, $A^\theta A$ is normal (in particular, we find that $A^\theta A = AA^\theta = -A - A^\theta$). By the spectral theorem, $A$ is unitarily diagonalizable. Because $A$ is diagonalizable with $0$ as its only eigenvalue, $A = 0$.

Since $A+I$ is unitary, $A$ is normal. Thus it is sufficient to prove that the only eigenvalue of $A$ is $0$.

Since $A+I$ is unitary, an eigenvalue of $A$ must be of the form $z-1$ where $|z|=1$. The corresponding eigenvalues of $A^2+I$ and $A^3+I$ are then $(z-1)^2+1$ and $(z-1)^3+1$. It can be seen that the only $z$ with $|z|=1$ and $|(z-1)^3+1|=1$ is $1$.