 Solution

Since $A$ is an invertible matrix, so is its Hermitian conjugate, andIf $A$ is a normal matrix, substitute $A^{†}A=A{A}^{†}$ into the second identity to obtain $B{B}^{†}=I$, so $B$ is a unitary matrix. Conversely, if $B$ is a unitary matrix, multiply $B{B}^{†}=I$ on the left by $A$ and on the right by $A^{†}$. This gives $A^{†}A=A{A}^{†}$, the defining normal matrix identity. Thus $B$ is a unitary matrix exactly when $A$ is a normal matrix.