Let $B$ and $C$ be complete Boolean algebras. To avoid triviality I may also want them to be atomless. For $b\in B$ nonzero, denote $B\upharpoonright b=\{p\in B:p\leq b\}$, which can be viewed as a complete Boolean algebra with maximal element $b$. By $B^+$ I mean the nonzero elements of $B$. The following is well-known, see e.g. this post by Hamkins.

TFAE:

(i) For every $C$-generic filter $H$, $V[H]$ contains a $B$-generic $G$ such that $V[G]=V[H]$, and vice versa.

(ii) $\{c\in C^+:\exists b\in B^+(C\upharpoonright c\simeq B\upharpoonright b)\}$ is dense in $C^+$, and vice versa.

Questions:

Can (ii) be strengthened to the following: there exist maximal antichains $A_1\subseteq B$ and $A_2\subseteq C$ such that for every $b\in A_1$, $B\upharpoonright b\simeq C\upharpoonright c$ for some $c\in A_2$, and vice versa.

Consider (iii) $B$ completely embeds into $C$ and vice versa. How is this related to (i) and (ii)? What is an example showing $B$ and $C$ may not be isomorphic?

The motivation for the first question is that $B$ and $C$ are not necessarily isomorphic, since for example $C$ could be a direct product of a large number of $B$'s, so they cannot be isomorphic for cardinality reason; direct product of Boolean algebras corresponds to disjoint sum (or lottery sum) of posets. I wonder if this is "the only obstacle" in some sense (if $B$ is a complete Boolean algebra and $A\subseteq B$ is a maximal antichain, then $B$ is isomorphic to $\prod_{a\in A}B\upharpoonright a$).