Let $\mathfrak{F}(0)$ be the set of all bijections $\mathbb{N}\mapsto\mathbb{N}$, and let $\mathfrak{F}(n+1)$ be the set of all bijections $\mathfrak{F}(n)\mapsto\mathfrak{F}(n)$. Given $\alpha\in\mathfrak{F}(n)$, is there a lower bound on the cardinality of $C_{\mathfrak{F}(n)}(\alpha)$, the centralizer of $\alpha$? Of course, since $C_{\mathfrak{F}(n)}(I)=\mathfrak{F}(n)$, where $I$ is the identity, the cardinality of the centralizer can be the same as that of $\mathfrak{F}(n)$. So the question is, is it possible to, using the axiom of choice as necessary, exhibit $\alpha\in\mathfrak{F}(n)$ such that $C_{\mathfrak{F}(n)}(\alpha)<\mathfrak{F}(n)$? Or, making things a little more concrete, is it possible to create $f:\mathbb{R}\mapsto\mathbb{R}$, a bijection, such that $C_{\mathfrak{F}(1)}(f)<2^{\mathbb{R}}$?
As far as I understand, the answer is negative (assuming $n>0$). It is easy to prove the following: Claim. Let $X$ be an uncountable set. Let $G$ be the group of all bijections $X\to X$. Then for any $g\in G$, its centralizer $Z(g)\subset G$ satisfies $Z(g)=G$. (This implies the claim, since all of $\mathfrak{F}(n)$ are uncountable.) Proof. It suffices to show that $Z(g)\geG$. Consider the partition of $X$ into orbits under the action of $g$. Each orbit is identified with a cyclic group ${\mathbb Z}/n{\mathbb Z}$ ($n\ge 0$) with $g$ acting as the shift by one. Since there are countably many isomorphism types of orbits, and each has countably many elements, there exists a particular $n\ge 0$ such that the corresponding set of orbits $X_n$ of this type has $X_n=X$. Now it is easy to see that the map from $Z(g)$ to the bijections of $X_n$ is surjective. (And hence the cardinality of $Z(g)$ is at least the cardinality of the group of bijections of $X_n$.) 

