SIDH: What if the two kernel generators are chosen in the same torsion group?

In SIDH, either party chooses its secret point $R_A = [m_A]P_A+[n_A]Q_A \in E[\ell_A^{e_A}]$, $R_B = [m_B]P_B+[n_B]Q_B \in E[\ell_B^{e_B}]$ from two different sets $E[\ell_A^{e_A}]$ and $E[\ell_B^{e_B}]$. What is the issue if the two points are chosen from the same set ($E[\ell_A^{e_A}]$ or $E[\ell_B^{e_B}]$)?

Very bad things happen. Let $E' := E/〈R_A〉$ be the image curve of the isogeny $φ:E→E'$ with kernel $R_A$:

1. $φ$ is not injective on $E[\ell_A^{e_A}]$: it maps it to a cyclic subgroup $G ⊂ E'[\ell_A^{e_A}]$;
2. The isogeny $\hat{φ}:E' → E'/G$ is the dual isogeny of $φ$;
3. In particular $E'/G$ is isomorphic to $E$.

Thus if both $R_A$ and $R_B$ were points in $E[\ell_A^{e_A}]$, necessarily $φ(R_B) ⊂ G$. If $φ(R_B)$ happens to be a generator of $G$ (the most likely case), then the shared "secret" of this variant of SIDH is the starting curve. In full generality, the shared secret would end up being one of the curves on Alice's path from $E$ to $E'$. In any case, this is clearly insecure.