Antichains and measure-preserving actions on Boolean algebras

Regarding the question is your last paragraph, there is the following often-studied but not-quite-equivalent-to-your property:

• A Boolean algebra $\mathbb{B}$ is almost homogeneous if for every nonzero $a,b\in\mathbb{B}$ there is an automorphism $\pi$ with $0\lt \pi(a)\wedge b$. Similarly, a group $G$ acts almost homogeneously if such $\pi$ can be found in $G$. This concept is also applied to partial orders in the natural way.

This is a weakening of your property, but it turns out to be critically important in the forcing context, since the theory forced by an almost homogeneous Boolean algebra does not depend on the generic filter.

I think the answer to first part of your question is affirmative. Begin by noting that the requirement on the automorphism $\Phi$ that all orbits are antichains implies that $\Phi^2$ is the identity. To see this suppose $\Phi$ is an automorphism acting on $B$ and let $a\in B$ be such that $\Phi(a) \neq a$. Since every orbit is an antichain, it must be that $a\cap \Phi(a) = \emptyset$. But now $\Phi(a\cup \Phi(a)) = \Phi(a)\cup \Phi(\Phi(a))$ and hence the orbit of $a\cup \Phi(a)$ is not an antichain unless $\Phi^2(a) = a$.

Now construct the probability measure $\mu$ by choosing $a_0\in B$ such that $\Phi(a_0)\neq a_0$ (and hence $\Phi(a_0)\cap a_0 = \emptyset$). It follows that if $b_0$ is the complement of $a_0\cup\Phi(a_0)$ then $\Phi(b_0) = b_0$ because $$\emptyset = \Phi(b_0\cap (a_0\cup\Phi(a_0))) = \Phi(b_0)\cap (\Phi(a_0)\cup\Phi^2(a_0)) = \Phi(b_0)\cap (\Phi(a_0)\cup a_0) $$ and hence $\Phi(b_0)\cap b_0\neq \emptyset$ --- so the antichain property yields that $\Phi(b_0) = b_0$. Let $\mu(a_0) = \mu(\Phi(a_0)) = 1/2$. So $\mu(b_0) = 0$ and this much of $\mu$ is preserved by $\Phi$.

Next choose a partition of $a_0$ into disjoint sets $a_{0,0} $ and $a_{0,1}$ and let $\mu(a_{0,i} )= \mu(\Phi(a_{0,i})) = 1/4$ Continuing in this spirit will define a probability measure on $B$ that is $\Phi$ invariant.

This is a follow-up to Juris’ answer, but it is a bit too long for a comment.

There is in fact no nontrivial group $G$ of automorphisms of $B$ such that all orbits are antichains, where $B$ is any Boolean algebra with more than $4$ elements.

Assume for contradiction that $f\in G$ is not the identity. If $a$ is any element of $B$ such that $f(a)\ne a$, then $a\land f(a)=0$ as the orbit of $a$ is an antichain; the orbit of $-a$ is also an antichain, hence $-a\land-f(a)=0$, which together imply $f(a)=-a$. Assume there is $b\in B$ such that $f(b)=b$ and $b\ne0,1$. Then $f(a\lor b)=-a\lor b$, but on the other hand we already know $f(a\lor b)\in\{a\lor b,-(a\lor b)\}$, which is easily seen to imply $b=1$ or $b=0$, a contradiction. Thus, $f(b)=-b$ for every $b\ne0,1$. However, this means that $f$ is order-reversing on $B\smallsetminus\{0,1\}$, whereas being an isomorphism, it is also order-preversing. This is a contradiction as long as we can find $0< a< b< 1$ in $B$, which we always can if $|B|>4$.