I am trying to prove the claim that:
If two polynomials satisfy $|P|^{2}=|Q|^{2}$, then they must have the same number of zeros, including the multiplicities.
It is immediate that they must have the same number of zeros, since if $P(\alpha)=0$ but $Q(\alpha)\neq 0$ for some $\alpha$, then $|P(\alpha)|^{2}=0\neq|Q(\alpha)|^{2}$, a contradiction.
The multiplicities are tricker, I had a proof but I don't know if it is correct:
If $P(z)=(z-\alpha)^{n}h_{1}(z)$ and $Q(z)=(z-\alpha)^{m}h_{2}(z)$ with $h_{1}(\alpha), h_{2}(\alpha)\neq 0$ holomorphic. WLOG assume $n=m+1$.
Then, $|P|^{2}=|Q|^{2}$ implies that $$|x-\alpha|^{2m+2}|h_{1}(x)|^{2}=|x-\alpha|^{2m}|h_{2}(x)|^{2},$$ so that $$|x-\alpha|^{2m}\Big(|x-\alpha|^{2}|h_{1}(x)|^{2}-|h_{2}(x)|^{2}\Big)=0.$$
This implies that, for $x\neq\alpha$, $$|h_{2}(x)|^{2}=|x-\alpha|^{2}|h_{1}(x)|^{2},$$ so that $$|q(x)|^{2}=|x-\alpha|^{2m+2}|h_{1}(x)|^{2},$$ so $$q(x)=\pm C(x-\alpha)^{m+1}h_{1}(x),$$ where $C$ is a constant of modulus $1$.
So the multiplicities are the same.
Is my proof correct? It seems really wired, but I don't know where I got wrong.
Thank you!
Edit 1:
I am editing this post to give a proof following the idea of "user8675309". In fact, in the problem I faced, $|P|^{2}=|Q|^{2}$ only on $\mathbb{S}^{1}$, but all the roots of $P$ and $Q$ lie on $\mathbb{S}^{1}$. This makes the thing a little bit more complicated. Below is the proof with this restricted condition:
$|P|^{2}=|Q|^{2}$ on $\mathbb{S}^{1}$ implies that there is some polynomial with real coefficients, say $G(z)$, is given by $P(z)$ times its conjugate or $Q(z)$ times its conjugate, for $z\in\mathbb{S}^{1}$. Since $G$ is the same polynomial, $P$ and $Q$ must have same zeros counting multiplicities on $\mathbb{S}^{1}$. But all of their zeros lie on $\mathbb{S}^{1}$, so we are done.
I will leave this post open for a few days, in case of further discussion or anyone post a solution so that I can accept it. Then I will answer my own post.
Since the discussion is finished, I am answering my own post to close it.
In fact, in the problem I faced, $|P|^{2}=|Q|^{2}$ only on $\mathbb{S}^{1}$, but all the roots of $P$ and $Q$ lie on $\mathbb{S}^{1}$. This makes the thing a little bit more complicated. Below is the proof with this restricted condition:
$|P|^{2}=|Q|^{2}$ on $\mathbb{S}^{1}$ implies that there is some polynomial with real coefficients, say $G(z)$, is given by $P(z)$ times its conjugate or $Q(z)$ times its conjugate, for $z\in\mathbb{S}^{1}$. Since $G$ is the same polynomial, $P$ and $Q$ must have same zeros counting multiplicities on $\mathbb{S}^{1}$. But all of their zeros lie on $\mathbb{S}^{1}$, so we are done.
I do appreciate user8675309, Gabriel and Conrad for their solution of general cases.