= Solution
In the <Rational map approximation for Skyrmions>, stereographic coordinate $z$ describes the direction $\widehat{\mathbf x}\in S^2$, and a degree-$B$ rational map $R(z)=p(z)/q(z)$ defines a unit vector $\mathbf n_R$. The <Skyrme model> field is approximated by
$$
U(r,z)=\exp[i f(r)\mathbf n_R(z)\mathbin\cdot\boldsymbol\tau],
\qquad f(0)=\pi,\quad f(\infty)=0.
$$
Its <baryon number> is the degree $B$ of $R$. Angular integration reduces the energy to a radial variational problem,
$$
E=4\pi\int_0^\infty\left[
r^2f'^2+2B(1+f'^2)\sin^2f
+\mathcal I\frac{\sin^4f}{r^2}\right]dr
$$
up to the conventional overall normalization, where the angular functional $\mathcal I$ depends only on $R$. One first minimizes $\mathcal I$ among degree-$B$ maps and then minimizes over the profile $f$. This efficiently captures the topology, energy, and polyhedral symmetries of many Skyrmions.
Let $\omega=e^{2\pi i/5}$. Since $\omega^5=1$,
$$
R(\omega z)=\omega^2R(z),
$$
which is a fivefold spatial rotation accompanied by a target-space rotation. The real coefficients also give $R(\bar z)=\overline{R(z)}$, while direct substitution gives
$$
R(-1/z)=-\frac1{R(z)}.
$$
Together these transformations extend the cyclic symmetry to the stated $D_{5d}$ symmetry.
For $p=z^7-7z^2$ and $q=7z^5+1$, the <Wronskian> is
$$
\boxed{W=p'q-pq'=14z(z^{10}+11z^5-1)}.
$$
Besides $z=0$, put $y=z^5$. Then
$$
y^2+11y-1=0,
\qquad
y_\pm=\frac{-11\pm5\sqrt5}{2}.
$$
Thus five zeros lie on the circle
$$
|z|=\left(\frac{5\sqrt5-11}{2}\right)^{1/5}
$$
at arguments $2\pi k/5$, and five lie on the reciprocal circle at arguments $(2k+1)\pi/5$. The polynomial has degree eleven, so the twelfth zero lies at $z=\infty$. On the Riemann sphere, the zeros therefore form two opposite poles and two staggered pentagonal rings: the twelve vertices of an icosahedron.
The angular baryon-density factor is proportional to $|dR/dz|^2$ and vanishes at these critical directions. The Wronskian zeros therefore point toward twelve holes in the baryon-density surface. They are the face centers of the dodecahedral $B=7$ Skyrmion, equivalently the vertices of its dual icosahedron, and make its icosahedral symmetry visible directly in the rational map.
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