Functions on a Boolean algebra that commute with every automorphism Is there a natural characterisation of the functions on a Boolean algebra that commute with all Boolean automorphisms of that algebra? In other words, the $f: B \to B$ such that $f \circ \pi = \pi \circ f$ for every automorphism $\pi$.
I'd be happy to know if there's a simple characterisation in the easiest case where the Boolean algebra is complete and atomic.
 A: I will describe how to construct all unary functions $f\colon B\to B$
that commute with the automorphisms of $B$.
Let $G = \textrm{Aut}(B)$ be the automorphism group of $G$.
For $a\in B$ let
$G_a$ be the stabilizer of $a$.
Let $B_a$ be the set of all fixed points of $G_a$.
If $f$ commutes with all automorphisms of $B$, then:
Claim 1. If $f(a)=b$, $a'=\pi a$, and $b'=\pi b$ for $\pi\in G$,
then $f(a')=b'$.
[I am just rewriting $f(\pi a)=\pi f(a)$.]
Claim 2. $f(a)\in B_a$.
[If $f(a)=b$ and $\pi a = a$, then $b=f(a)=f(\pi a) =
\pi f(a) = \pi b$, so $\pi b =b$. Thus $b\in B_a$.]
Claim 3. $f$ is arbitrary subject to the restrictions in Claims 1 and 2.
[Claim 1 expresses that the graph of $f$ is a $G$-invariant
binary relation, while Claim 2 is the condition necessary
for this to be the graph of a function.]
The claims show that determining the possible $f$'s
reduces to determining the $G$-orbits of $B$
and also determining $B_a$
for each $a$. One then simply selects one element $a$ from each $G$-orbit,
defines $f(a)$ arbitrarily subject to $f(a)\in B_a$, 
then propagates this partial description of $f$ in
the unique way allowable by Claim 1 to make it a total function.
There exist rigid Boolean algebras ($G=\{1\}$),
and for such algebras any function $f\colon B\to B$
commutes with all automorphisms.
If $B = {\mathcal P}(X)$,
then the orbit of an element $a\in {\mathcal P}(X)$ is determined
by the pair of cardinals $(|a|, |X-a|)$: two elements belong
to the same $G$-orbit iff they have the same `cardinality type',
$(|a|, |X-a|)$. Also, if $a\in {\mathcal P}(X)$, then
$B_a = \{\emptyset, a, X-a, X\}$, so $f(a)$ must be one
of these elements. Thus, in this case, to determine
a possible $f$,
one should select a cardinality type, then decide which case


*

* $f(a)=\emptyset$, 

* $f(a)=a$, 

* $f(a)=X-a$, or

* $f(a)=X$ 


holds uniformly for all elements $a$ of this cardinality type. Then do
the same for every cardinality type.
