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 $|Z(g)|\ge|G|$. 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 $Z(g)$ acts transitively on $X_n$. (And hence its cardinality is at least the cardinality of the group of bijections of $X_n$.)

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 $|Z(g)|\ge|G|$. 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 $X_n$.)