Solution: Given that $A \in M_n(\mathbb{C} )$ is a normal matrix. We claim that if all the eigenvalues of $A$ are real, then $A$ is Hermitian, that is, $A^{\ast} = A$. We recall that the Spectral theorem states that a matrix $A$ is normal if and only if it is unitarily similar to a diagonal matrix, that is, there exists a unitary matrix $U$ such that \[ A = U D U^{\ast} \] where $D$ is a diagonal matrix containing the eigenvalues of $A$. Since the eigenvalues of $A$ are real, \begin{align*} A^{\ast} = \left( U D U^{\ast} \right) ^{\ast} = U D^{\ast} U^{\ast} = U D U^{\ast} = A. \end{align*} Thus, $A$ is Hermitian.

Solution: Let us now consider the statement II. We claim that if all the eigenvalues of $A$ have absolute value $1$, then $A$ is unitary, that is, $A A^{\ast} = I = A^{\ast} A$. We will again use the spectral theorem. There exists a unitary matrix $U$ such that \begin{align*} A = U D U^{\ast} \implies A^{\ast} = U D^{\ast} U^{\ast}. \end{align*} So, we have \begin{align*} A A^{\ast} & = \left( U D U^{\ast} \right) \left( U D^{\ast} U^{\ast} \right) = U D (U^{\ast} U) D^{\ast} U^{\ast} \\ & = UD D^{\ast}(I) U^{\ast} = UD D^{\ast} U^{\ast}. \end{align*} If $D = \mathrm{diag} \begin{bmatrix} \lambda_1 & \lambda_2 & \cdots & \lambda_n \\ \end{bmatrix} $, then \[ D D^{\ast} = \mathrm{diag} \begin{bmatrix} \lambda _1 & \lambda _2 & \cdots & \lambda _4 \\ \end{bmatrix} \begin{bmatrix} \overline{\lambda}_1 & \overline{\lambda }_2 & \cdots & \overline{\lambda }_n \end{bmatrix} = I. \] The last equality is due to the hypothesis that all the eigenvalues have absolute value $1$. Thus, \begin{align*} A A^{\ast} = U DD^{\ast} U^{\ast} = U I U^{\ast} = UU^{\ast} = I. \end{align*} Similarly, $A^{\ast} A = I$ and hence $A$ is unitary matrix.