= Solution
Since $A$ is an <invertible matrix>, so is its <Hermitian conjugate>, and
$$
B^\dagger=A(A^\dagger)^{-1},\qquad BB^\dagger=A^{-1}A^\dagger A(A^\dagger)^{-1}.
$$
If $A$ is a <normal matrix>, substitute $A^\dagger A=AA^\dagger$ into the second identity to obtain $BB^\dagger=I$, so $B$ is a <unitary matrix>. Conversely, if $B$ is a <unitary matrix>, multiply $BB^\dagger=I$ on the left by $A$ and on the right by $A^\dagger$. This gives $A^\dagger A=AA^\dagger$, the defining <normal matrix> identity. Thus \b[$B$ is a <unitary matrix> exactly when $A$ is a <normal matrix>].
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