In a comment on one of my answers, I claimed that the abelian group generated by a set of $S$ generators, each of order two, could take on any infinite cardinality; this is equivalent to saying that, if we let $\mathscr P'(S)$ be the set of finite subsets of $S$ (with with the aforementioned group is equinumerous), then, for any infinite cardinality $A$, there is a set of the form $\mathscr P'(S)$ with cardinality $A$.
The axiom of choice clearly implies that this is true, because it implies that $A$ and $\mathscr P'(A)$ are equinumerous. (And that this holds for all $A$ is clearly equivalent to AC, since it implies that $A\times A$ and $A$ are also equinumerous). However, when we don't have the axiom of choice, is it still provable that, even if $\mathscr P'(A)$ and $A$ are not equinumerous, there still must exist some $S$ such that $\mathscr P'(S)$ and $A$ are equinumerous?
Suppose that $\varphi:A\to\wp'(S)$ is a bijection. Let $A_0=\{a\in A:|\varphi(a)|\text{ is even}\}$, and let $A_1=A\setminus A_0$. Then $\{A_0,A_1\}$ is a partition of $A$ into infinite sets, and $A$ is therefore not amorphous. However, there are models of $\mathsf{ZF+\neg AC}$ in which there are amorphous sets.