= Solution
For every $x$,
$$
\|\alpha x\|^2=\langle x,\alpha^*\alpha x\rangle,
\qquad
\|\alpha^*x\|^2=\langle x,\alpha\alpha^*x\rangle.
$$
If $\alpha$ is normal, these quantities are equal. Conversely, equality of the norms for all $x$ gives
$$
\langle x,(\alpha^*\alpha-\alpha\alpha^*)x\rangle=0.
$$
The operator in parentheses is self-adjoint, so polarization makes it zero. Therefore
$$
\boxed{\alpha\alpha^*=\alpha^*\alpha
\iff \|\alpha x\|=\|\alpha^*x\|\quad\hbox{for all }x}.
$$
The same equivalence holds over a real <inner product> space: for a self-adjoint operator $T$, the real polarization identity
$$
4\langle Tx,y\rangle
=\langle T(x+y),x+y\rangle
-\langle T(x-y),x-y\rangle
$$
shows that a vanishing quadratic form forces $T=0$.
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