Degrees of the two Rayleigh–Ritz discriminant parts for rational normal curves

For $d \ge 2$ let $$C_d = \{[a^d : a^{d-1}b : \cdots : b^d] : [a : b] \in \mathbb{P}^1\} \subset \mathbb{P}^d$$ be the rational normal curve, in this unweighted embedding. For a complex symmetric matrix $H$ of order $d + 1$, let $\rho_H(z) = z^{\mathsf T} H z / z^{\mathsf T} z$ on the part of $C_d$ where $z^{\mathsf T} z \ne 0$; transposes do not conjugate. Let $\mathcal{R}$ be the Zariski closure in $C_d \times \mathbb{P}(\operatorname{Sym}_{d+1}(\mathbb{C}))$ of the pairs $([z], [H])$ for which $[z]$ is a critical point of $\rho_H$ on $C_d$, choose homogeneous generators $g_i$ of its ideal, and let $$\mathcal{B} = \{([z], [H]) \in \mathcal{R} : \operatorname{rank}(\partial g_i / \partial z_j) \le d - 1\}.$$ With $Q = \{([z], [H]) : z^{\mathsf T} z = 0\}$ and $\pi$ the projection to the matrix factor, the isotropic and nonisotropic parts of the Rayleigh–Ritz discriminant are $$\Sigma_{\mathrm{iso}} = \pi(\mathcal{B} \cap Q), \qquad \Sigma_{\mathrm{noniso}} = \overline{\pi(\mathcal{B} \setminus Q)},$$ the closure being Zariski. Degrees are those of the reduced algebraic sets, all components included. Borovik, Friedman, Hoşten and Pfeffer showed that the whole discriminant $\Sigma_{\mathrm{iso}} \cup \Sigma_{\mathrm{noniso}}$ is the zero set of the discriminant of a binary form of degree $2(2d - 1)$ whose coefficients are linear in $H$, a polynomial of degree $2(4d - 3)$. For $d = 2$ and $d = 3$ their symbolic computations give the degrees below, and they report numerical checks for $d \le 10$.

Problem. Are $\Sigma_{\mathrm{noniso}}$ and $\Sigma_{\mathrm{iso}}$ hypersurfaces with $$\deg \Sigma_{\mathrm{noniso}} = 6(d - 1) \quad \text{and} \quad \deg \Sigma_{\mathrm{iso}} = 2d$$ for every $d \ge 2$?