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A forcing construction I'm trying to do seems to require a complete atomless boolean algebra (used as a forcing poset) that is "semi-rigid" in the sense defined below. I'm wondering if anyone has studied this property, maybe using a different term, and shown it to be realizable or not realizable.

A rigid boolean algebra is one that has no nontrivial automorphisms. Call a boolean algebra "semi-rigid" if it does have at least one nontrivial automorphism, but none of these has any fixed points other than 0 and 1 (the algebra's least and greatest elements). The justification for calling this "semi-rigidity" is that although there are nontrivial automorphisms, knowing where an automorphism puts one element -- any element, other than 0 and 1 -- completely determines where it will put every element; there is no further flexibility. (For if $\phi$ and $\phi'$ were distinct automorphisms that mapped some $b$ ($\neq 0,1$) to the same element, then $\phi^{-1} \circ \phi'$ would be a nontrivial automorphism with fixed point $b$.) The four-element algebra $\{0, b, \neg b, 1\}$ is a simple example of semi-rigidity.

Preliminary question: Are there semi-rigid complete atomless boolean algebras (CABA's)? I suspect you can get one by gluing together copies of a rigid CABA $C$ (such $C$'s are well known to exist): let $B$ be a CABA with an element $b$ such that $B \upharpoonright b$ and $B \upharpoonright \neg b$ (i.e. the principal ideals with greatest elements $b$ and $\neg b$ respectively) are isomorphic copies of $C$. I suspect this $B$ will have exactly one nontrivial automorphism, which interchanges $B \upharpoonright b$ and $B \upharpoonright \neg b$.

Even if this is right and such a $B$ can be proved semi-rigid, though, it would be an uninteresting example because its semi-rigidity would reduce to rigidity (of principal ideals). So my main question is:

Question: Are there semi-rigid CABA's none of whose principal ideals are rigid?

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