Square ICA is identifiable up to signs and permutation with at most one Gaussian source

Work on the probability space of $S$ and define $T=A^{\mathrm{\prime }-1}AS$. The given equality implies by applying the fixed invertible linear map $A^{\mathrm{\prime }-1}$ to both laws. In particular the coordinates of $T$ are mutually independent, centered and have unit variance. Write $O=A^{\mathrm{\prime }-1}A$, so $T=OS$ on this probability space. Independence and centering give $\mathrm{Cov}(S)=\mathrm{Cov}(T)=I_p$, and covariance under this linear map yields $O{O}^T=I_p$. Since $O$ is square and invertible, also $O^{T}O=I_p$; it is orthogonal. This covariance calculation is the relative whitening reduction and requires neither a density nor moments beyond the given second moments.

By [F2], every non-Gaussian column of $O$ is a signed coordinate vector. Distinct such columns occupy distinct rows, because they are orthogonal. There are at least $p-1$ such columns. If there are $p$ they occupy all rows and $O$ is already a signed-permutation matrix. Otherwise there are exactly $p-1$, leaving one row $r$ and the one possible Gaussian column $i$. Its inner product with each occupied signed coordinate column is zero, so $O_{ji}=0$ on every occupied row $j$. Its norm is one, so $O_{ri}=1$ or $-1$. Thus $O$ is a signed-permutation matrix also in this case. For $p=1$ the same argument has zero occupied rows and its single norm-one column is $1$ or $-1$.

Put $Q=O^T$, again a signed-permutation matrix. From $O=A^{\mathrm{\prime }-1}A$ and $O^{-1}=O^T$ we get $A^{\mathrm{\prime }}=A{O}^T=AQ$. Also . These are exactly the asserted matrix and source-law conclusions.