Fix an integer $n\ge1$, the order of the symmetry. In this section a \emph{pair} $(W,B)$ satisfies (M1) and (M2) of
Definition~\ref{def:model}. We use the
criterion of Proposition~\ref{prop:criterion} with base point $w_0=0$, so that $v_W=W/(wW')$ and $\sigma_B=wB'/B$ on the
circle. A pair is \emph{$n$-fold symmetric}\label{term:symmetric} if $W(e^{2\pi i/n} w)=e^{2\pi i/n}W(w)$ and the zero set of $B$, counted with multiplicities,
is invariant under $w\mapsto e^{2\pi i/n} w$. By the uniqueness of the normalised Riemann map, $W(e^{2\pi i/n} w)=e^{2\pi i/n}W(w)$ holds if and
only if $W(\mathbb D)$ is invariant under the rotation by $2\pi/n$.

\subsection{The root transform}
Let $\Psi\in\mathcal S$ be holomorphic and univalent on $\{|u|<R\}$, $R>1$. Then $\Psi(u)/u$ is holomorphic there, equal to $1$ at
$0$, and zero-free, because $\Psi$ is injective and $\Psi(0)=0$; so $\hat\Psi=(\Psi(u)/u)^{1/n}$, the branch with $\hat\Psi(0)=1$, exists and has
no zero. Put
\[
  \Psi_n(w)=w\,\hat\Psi(w^n)\qquad(|w|<R^{1/n}),\qquad\text{and}\qquad B(w)=\widetilde B(w^n)
\]
for a finite Blaschke product $\widetilde B$. This is the $n$-th root transform of~\cite[Rem.~8.6]{JatarNg2026b}, where its
admissibility condition is given in the open-disk setting. We record the closed-disk form in the notation of
Proposition~\ref{prop:criterion}.

\begin{lemma}[{\cite[Rem.~8.6]{JatarNg2026b}}]\label{lem:rt}
\textup{(a)} $\Psi_n$ satisfies \textup{(M1)} with radius $R^{1/n}>1$, and $\Psi_n(e^{2\pi i/n} w)=e^{2\pi i/n}\Psi_n(w)$.
\textup{(b)} $B$ is a finite Blaschke product of degree $n\deg\widetilde B$, and $B(0)=\widetilde B(0)$.
\textup{(c)} $(\Psi_n,B)$ is a Blaschke model if and only if $\widetilde\gamma:=1+n\,\sigma_{\widetilde B}\,v_\Psi$ has no zero on the circle
and $\operatorname{wind}(\widetilde\gamma)=-\deg\widetilde B$.
\end{lemma}

\begin{proof}
(a) $\Psi_n(e^{2\pi i/n} w)=e^{2\pi i/n}\Psi_n(w)$ and $\Psi_n^n=\Psi(w^n)$. If $\Psi_n(w_1)=\Psi_n(w_2)$, then $\Psi(w_1^n)=\Psi(w_2^n)$, so
$w_1^n=w_2^n$ and $w_2=e^{2\pi ij/n}w_1$; then $\Psi_n(w_1)=e^{2\pi ij/n}\Psi_n(w_1)$, so $e^{2\pi ij/n}=1$ or $\Psi_n(w_1)=0$, that is $w_1=0=w_2$.
Also $\Psi_n'(0)=\hat\Psi(0)=1$.
(b) For a zero $b\neq0$ of $\widetilde B$ and $\beta^n=b$: $\prod_{j<n}(w-e^{2\pi ij/n}\beta)=w^n-b$ and
$\prod_{j<n}(1-\overline{e^{2\pi ij/n}\beta}\,w)=1-\bar b\,w^n$; for $b=0$ the factor becomes $w^n$.
(c) The logarithmic derivative of $\Psi_n^n=\Psi(w^n)$ gives $v_{\Psi_n}(w)=v_\Psi(w^n)$, and
$\sigma_B(w)=n\,\sigma_{\widetilde B}(w^n)$. Hence $\gamma=1+\sigma_Bv_{\Psi_n}=\widetilde\gamma(w^n)$, which vanishes on the circle
if and only if $\widetilde\gamma$ does, and $\operatorname{wind}(\gamma)=n\operatorname{wind}(\widetilde\gamma)$. By
Proposition~\ref{prop:criterion}, (M3) means $\operatorname{wind}(\gamma)=-\deg B=-n\deg\widetilde B$.
\end{proof}

In the open-disk form of~\cite[Rem.~8.6]{JatarNg2026b} the zeros of $\widetilde B$ must also be simple. In (c) no such condition is
needed: a multiple zero of $\widetilde B$ is a zero of $(\Psi_nB)'$ in $\mathbb D$, and the winding count of
Proposition~\ref{prop:criterion} includes it.

\subsection{Every symmetric pair is a root transform}

Jatar and Ng note this correspondence for polynomials of the form $w\,p(w^n)$~\cite[Rem.~8.6]{JatarNg2026b}. We need it at the level
of pairs, where the hypothesis $B(0)\neq0$ enters.

\begin{theorem}\label{thm:rtconverse}
Let $(W,B)$ be an $n$-fold symmetric pair with $B(0)\neq0$. Then $W=\Psi_n$ and $B=\widetilde B(w^n)$ for some $\Psi$ as above
(with radius $R^n$) and some finite Blaschke product $\widetilde B$.
\end{theorem}

\begin{proof}
Let $W$ be univalent on $\{|w|<R\}$. By the identity theorem $W(e^{2\pi i/n} w)=e^{2\pi i/n}W(w)$ there, so $W^n$ is invariant under
$w\mapsto e^{2\pi i/n} w$ and its Taylor series contains only the powers $w^{nm}$: $W^n=\Psi(w^n)$ with $\Psi$ holomorphic on
$\{|u|<R^n\}$ and $\Psi(u)=u+\cdots$. $\Psi$ is injective: if $\Psi(u_1)=\Psi(u_2)$, choose $w_i^n=u_i$; then
$W(w_1)^n=W(w_2)^n$, so $W(w_1)=e^{2\pi ij/n}W(w_2)=W(e^{2\pi ij/n}w_2)$, hence $w_1=e^{2\pi ij/n}w_2$ and $u_1=u_2$. Now
$\Psi_n^n=W^n$, and $\Psi_n$, $W$ have a simple zero at $0$ with derivative $1$ and no other zero, so $\Psi_n/W$ is
holomorphic and zero-free, with $n$-th power $1$ and value $1$ at $0$; it is the constant $1$.

Since $B(0)\neq0$, every orbit $\{e^{2\pi ij/n}\beta\}$ of a zero of $B$ has exactly $n$ points, all of the same multiplicity. Let
$\widetilde B$ have one zero $\beta^n$ per orbit, with that multiplicity. By (b) of Lemma~\ref{lem:rt}, $B$ and $\widetilde B(w^n)$ have
the same zeros with multiplicities, so their quotient is a unimodular constant, which we absorb into $\widetilde B$.
\end{proof}

The hypothesis $B(0)\neq0$ is needed: for $n=2$ the zero set $\{0\}$ of $B(w)=w$ is invariant, but $B$ is not of the form
$\widetilde B(w^2)$. It holds for every Blaschke model, since $(W B)'(0)=B(0)$.

\begin{corollary}\label{cor:rt}
For $\Psi$ as above and $t>0$ call a finite Blaschke product $\widetilde B$ \emph{$t$-allowed} for $\Psi$ if
$1+t\,\sigma_{\widetilde B}\,v_\Psi$ has no zero on the circle and winds $-\deg\widetilde B$ times, and let $\operatorname{gain}_t(\Psi)$ be the supremum
of $1/|\widetilde B(0)|$ over the $t$-allowed $\widetilde B$ $(\deg\widetilde B=0$ gives $\operatorname{gain}_t(\Psi)\ge1)$. Then for every integer
$n\ge1$:
\textup{(i)} $\operatorname{gain}_n(\Psi)\le K^*$;
\textup{(ii)} the supremum of the gains of $n$-fold symmetric Blaschke models equals $\sup_\Psi\operatorname{gain}_n(\Psi)$.
\end{corollary}

\begin{proof}
(i) By Lemma~\ref{lem:rt}(c) each $n$-allowed $\widetilde B$ gives the Blaschke model $(\Psi_n,\widetilde B(w^n))$ with the same gain;
apply Theorem~\ref{thm:BF}. (ii) A symmetric Blaschke model has $B(0)\neq0$, so by Theorem~\ref{thm:rtconverse} it is
$(\Psi_n,\widetilde B(w^n))$, and by Lemma~\ref{lem:rt}(c) its $\widetilde B$ is $n$-allowed for $\Psi$, with the same gain.
Conversely, for an $n$-allowed $\widetilde B$ the pair $(\Psi_n,\widetilde B(w^n))$ is a symmetric Blaschke model, since
$B(e^{2\pi i/n} w)=B(w)$.
\end{proof}

\begin{remark}\label{rem:rtpockets}
In the language of Section~\ref{sec:pockets}, the pockets of $\Psi_n$ are the $n$ preimages under $w\mapsto w^n$ of those of $\Psi$,
with the same depth and the same sign ($\theta\mapsto n\theta$ is increasing), and the condition of Lemma~\ref{lem:rt}(c) is
Theorem~\ref{thm:pocket} for $\Psi$ with $\hat\sigma=n\,\sigma_{\widetilde B}$. So a pocket of depth $x$ is on if and only if
$n\,x\,\sigma_{\widetilde B}>1$ there: for the quotient $\widetilde B$, symmetry by $2\pi/n$ acts as if every depth were multiplied by $n$. When Proposition~\ref{prop:guard} or
Lemma~\ref{lem:independence} is applied to the quotient, $\deg\widetilde B$ and $n$ times the Poisson kernels of the zeros of
$\widetilde B$ take the places of $\deg B$ and of the kernels.
\end{remark}