= Solution
For coprime polynomials $p,q$, the algebraic degree of $R=p/q:S^2\to S^2$ is
$$
\deg R=\max(\deg p,\deg q),
$$
and its Wronskian is $W=p'q-pq'$. Thus
$$
R=z^2:\quad \deg R=2,\quad W=2z,
$$
while for
$$
R=\frac{\sqrt3iz^2-1}{z^3-\sqrt3iz}
$$
one obtains
$$
\boxed{\deg R=3,\qquad W=-\sqrt3iz^4+6z^2-\sqrt3i}.
$$
For $R=z^2$, a domain rotation $z\mapsto e^{i\alpha}z$ is accompanied by a target rotation $R\mapsto e^{2i\alpha}R$. There is also a half-turn about every axis in the equatorial plane, with the corresponding target half-turn. These generate the axial dihedral rotational symmetry: the distinguished spatial axis is the $z$-axis, and the continuous rotations about it are accompanied by twice the angle in target space.
The <Rational map approximation for Skyrmions> uses
$$
U(r,z)=\exp\bigl(if(r)\,\widehat n_R(z)\cdot\boldsymbol\tau\bigr),
\qquad f(0)=\pi,\quad f(\infty)=0.
$$
The rational-map degree is the <Skyrmion> baryon number, while zeros of the Wronskian identify directions where the angular baryon density vanishes and therefore influence the shape and energy. For $R=z^2$ the result is the $B=2$ toroidal Skyrmion.
Quantization treats spatial and isospin rotations as collective coordinates and imposes the <Finkelstein-Rubinstein constraints>. For even baryon number, spin and isospin are integral. The axial constraint is $2J_3+I_3=0$, and the equatorial half-turn imposes $(-1)^{J+I}=-1$. Restricting to $J,I\leq1$ leaves
$$
\boxed{(J,I)=(1,0)\quad\text{and}\quad(J,I)=(0,1)}.
$$
The $(0,0)$ and $(1,1)$ representations violate the discrete constraint.
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