symmetric subspace of $\mathcal{H}_d^{\otimes 2}$ and in the third step we employed the fact that the SICs form a minimal $2$-design [@Renes:2004]. This shows that the modulus of $\Phi$ is equal to $\Phi_{\rm SIC}$ and thus the singular values of $\Phi$ and $\Phi_\text{SIC}$ coincide.

On the other hand, suppose $s(\Phi)=s(\Phi_\text{SIC})$. The product of all the singular values is the absolute value of the determinant [@Horn:1994], so $|\det\Phi^{-1}|=\det\Phi^{-1}_\text{SIC}\implies|\det\alpha|=1\implies\det G=\det G_{\rm SIC}\iff \{E_i\}$ is a SIC. Carrying through the consequences of the MIC being a SIC allows us to see from Eq.  that $\sigma_j$ is rank-$1$ because the upper bound is saturated for SIC probability vectors. We may expand the $\{\sigma_j\}$ in the SIC projector basis, $$\sigma_j=\sum_k\left[\beta\right]_{jk}\Pi_k\;.$$ Acting on both sides by a SIC POVM element and computing the trace of both sides, we see $$\left[\Phi^{-1}\right]_{ij}=\tr E_i\sigma_j=\sum_k\left[\beta\right]_{jk}\tr E_i\Pi_k=[\Phi^{-1}_{\rm SIC}\beta^T]_{ij}\;,$$ so $|\det\Phi^{-1}|=|\det\Phi^{-1}_{\rm SIC}||\det\beta^T|=\det\Phi^{-1}_{\rm SIC}$ implies $|\det\beta|=1$. Denoting the Gram matrix of states by $g$, we have, in the same way as before, $$\det g=(\det\beta)^2\det g_{\rm SIC}=\det g_{\rm SIC}\;.$$ We now prove that $\det g=\det g_{\rm SIC}$ implies that the basis of projectors forms a SIC. The following lemma is due to Huangjun Zhu [@Zhu:2017]. We only use part of Zhu’s conclusion, but the 