$S^3 \times S^5$ has isometry group $SO_4(\mathbb{R}) \times SO_6(\mathbb{R})$, which has $SU(2) \times SU(2) \times SU(4)$ as a four-fold cover. Since it appears that you aren't worrying too much about central terms, we can replace $S^3$ with $\mathbb{R}P^3$, or $S^5$ with $\mathbb{R}P^5$.\mathbb{R}P^5$, or take a quotient by a diagonal group of order 2. I'm pretty sure these are the only connected choices, because we can characterize homogeneous orbits by the stabilizers of points. In this case, you need a closed subgroup of Pati-Salam of dimension at least 13 whose intersection with each factor group is not the whole factor. There just aren't that many subgroups of suitably large dimension: we need a diagonally embedded$SU(2)$(possibly with a central translate) to get dimension at least 3 in the first factors, and we need Spin(5) in the last factor to get dimension at least 10. This forces the orbits to be connected components. 1$S^3 \times S^5$has isometry group$SO_4(\mathbb{R}) \times SO_6(\mathbb{R})$, which has$SU(2) \times SU(2) \times SU(4)$as a four-fold cover. Since it appears that you aren't worrying too much about central terms, we can replace$S^3$with$\mathbb{R}P^3$, or$S^5$with$\mathbb{R}P^5$. I'm pretty sure these are the only connected choices, because we can characterize homogeneous orbits by the stabilizers of points. In this case, you need a closed subgroup of Pati-Salam of dimension at least 13 whose intersection with each factor group is not the whole factor. There just aren't that many subgroups of suitably large dimension: we need a diagonally embedded$SU(2)\$ (possibly with a central translate) to get dimension at least 3 in the first factors, and we need Spin(5) in the last factor to get dimension at least 10. This forces the orbits to be connected components.