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A boolean algebra is rigid if it has no nontrivial automorphisms. Call it semi-rigid if none of its nontrivial automorphisms has any fixed points other than 0 and 1.* The four-element algebra $\{0, b, \neg b, 1\}$ is a simple example of a semi-rigidity. Preliminary question: Are there semi-rigid complete atomless boolean algebras (CABA's)? I suspect so: Let $B$ be a CABA with an element $b$ such that the principal ideals $B \upharpoonright b$ and $B \upharpoonright \neg b$ (having greatest elements $b$ and $\neg b$) are isomorphic copies of the same rigid CABA $C$. I suspect this $B$ has exactly one nontrivial automorphism, which interchanges $B \upharpoonright b$ and $B \upharpoonright \neg b$.

Even if this $B$ 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 question is: Are there semi-rigid CABA's none of whose principal ideals are rigid?

*The rationale for the term "semi-rigid" is this: Knowing where an automorphism of a semi-rigid boolean algebra maps one element -- any element, other than 0 and 1 -- determines where it maps all other elements; there is no further flexibility. For if $\phi$ and $\phi'$ were distinct automorphisms that mapped some $b$ to the same element, then $\phi^{-1} \circ \phi'$ would be a nontrivial automorphism with fixed point $b$.

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There is no such algebra. In fact, suppose that $B$ satisfies the indicated condition. Choose $a$ in $B$ with $a$ not equal to $0$ or $1$. Let $f$ be a nontrivial automorphism of the principal ideal determined by $a$. Define $g(x)=f(x \wedge a)\vee(x \wedge-a)$. Then $g$ is a nontrivial automorphism of $B$ with fixed point $-a$.

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