An equivalence relation on the space of polynomials in one complex variable

Question

Let $P(z)$ be a polynomial with complex variable $z$. We consider the following distribution for the roots of $P(z)=0$: the distribution is a triple $(n_{1},n_{2},n_{3})$ where these integers are the number of roots in the interior of unite circle, on the unit circle and out of unit disc, respectively.

$\ell^{2}$ is the Hilbert space of all square sumable sequence of complex numbers. $S_1$ is the shift operator on $\ell^{2}.$

The equivalence relation: We say two polynomials $P,Q$ are equivalent if $P(S_1)$ is conjugate to $Q(S_1)$ via an invertible operator in $B(\ell^{2})$.

Assume that $P,Q$ are two equivalent polynomials:

1.Must they have the same degree?

2.Must they have the same root distribution?

The motivation for the second question is that when $P,Q$ have the same degree and $P$ has $(n_{1},0,n_{3})$ distribution, then $Q$ has the same distribution as $P$. The reason is that $P(S_1)$ is a Fredhom operator of index $-n_1$. Obviously this property is invariant under conjucacy. So $Q$ has the same root distribution.

Answer 1

Much more is true. According to the answer by Alexandre Eremenko (I do not have this paper, but I completely trust him), from $P(\mathbb{S}^1)=Q(\mathbb{S}^1)$ (which are essential spectra of $P(T)$, $Q(T)$ as noted in the comments) it follows that $P=f(z^n)$, $Q=f(wz^m)$ for some number $w$, $|w|=1$, positive integers $n,m$ and polynomial $f$. Further, I claim that $n=m$, it would mean just $Q(z)=P(wz)$. Indeed, choose complex number $a$, $|a|<1$, so that $f(a)\notin P(\mathbb{S}^1)$, and consider polynomials $P(z)-f(a)$, $Q(z)-f(a)$. They do not have roots on $\mathbb{S}^1$, thus the numbers of their roots inside the unit disc must be the same (as noted in your post, these numbers are just Fredholm indices of $P(T)-f(a)$, $Q(t)-f(a)$). But these numbers equal $n\cdot k$, $m\cdot k$ resp., where $k>0$ is the number of roots of $f(z)-f(a)$ in the open unit disc. Thus $n=m$ as desired.