Consider structures $(A,f)$ encoding a Boolean algebra $A$ endowed with an automorphism $f$. There is an obvious notion of isomorphism between such structures.

Consider the endomorphism $\hat{\Phi}$ of the Boolean algebra $2^\omega$ of subsets of $\omega$ given by $A\mapsto \{a\in\omega:a+1\in A\}$. It induces an automorphism $\Phi$ of the quotient Boolean algebra $2^\omega/\mathrm{fin}$, where $\mathrm{fin}$ is the ideal of finite subsets. (Under Stone duality, this corresponds to the self-homeomorphism of the Stone-Čech remainder of $\omega$ induced by $n\mapsto n+1$.)

Whether $(2^\omega/\mathrm{fin},\Phi)$ and $(2^\omega/\mathrm{fin},\Phi^{-1})$ are isomorphic is essentially unknown (see my previous related question for more details).

My question is

Are $(2^\omega/\mathrm{fin},\Phi)$ and $(2^\omega/\mathrm{fin},\Phi^{-1})$ elementary equivalent?

That is, do they satisfy the same first-order sentences (in the language of Boolean algebras endowed with an automorphism)?

Note that $\Phi^n$ has exactly $2^{|n|}$ fixed points for $n\neq 0$. In particular, $(2^\omega/\mathrm{fin},\Phi)$ and $(2^\omega/\mathrm{fin},\Phi^n)$ are not not equivalent for $|n|\ge 2$.

If the question has a positive answer, it is tempting to ask whether one can characterize simply those $(A,f)$ having the same first-order theory as $(2^\omega/\mathrm{fin},\Phi)$.

2more comments