In ZFC, it is easy to prove there are more cardinals than elements of any set. Specifically, given a set $X$, pick a well-ordering of $X$, and then you can inject $X$ into the cardinals by mapping its $\alpha$th element to $\aleph_\alpha$.
My question is whether this can be proved in ZF (where "cardinal" is interpreted in the appropriate way in the absence of choice, not restricted to well-ordered cardinalities). To be precise, can ZF prove the following?
For any set $X$, there exists a function $f$ on $X$ such that for any distinct $x,y\in X$, there is no bijection between $f(x)$ and $f(y)$.
I imagine the answer is no but don't really know anything about how you could prove such a global statement about the cardinals in a model of ZF in order to get a counterexample.
At least ZFA cannot prove this assertion. The model constructed here should be the standard model of ZFA where choice for families of sets of size 2 fails.
Start with a model of ZFA with a countable set $A$ of atoms. Label the elements of $A$ as $a_n^i$ with $n<\omega$ and $i<2$. Think of $a_n^0, a_n^1$ of pairs and the set $A$ as an infinite set of these pairs. Let $G$ be the group of permutations $\pi$ of $A$ that maybe swap pairs, but do not do anything else, i.e. $\pi(a_n^i)\in\{a_n^i, a_n^{1-i}\}$ for all $n<\omega, i<2$. Let $\mathcal F$ be the filter generated by $\mathrm{fix}_G(E)$, where $E$ ranges over the finite subsets of $A$. This induces a permutation model $\mathcal V$ of hereditarily symmetric sets. Note that the group $G$ is commutative. This has the following consequence: As usual, any $\pi\in G$ exptends to an automorphism $\pi^+$ of $\mathcal V$. If $X\in\mathcal V$ then $\pi^+\upharpoonright X:X\rightarrow\pi^+(X)$ is a bijection that lies in $\mathcal V$. The reason being that $\pi^+$ commutes with any other $\mu^+$, $\mu\in G$, which is enough to see that $\pi^+\upharpoonright X$ is (hereditarily) symmetric.
If $f:A\rightarrow\mathcal V$ is a function in $\mathcal V$, then for cofinitely many $n$, $f(a_n^0)$ and $f(a_n^1)$ cannot be significantly different (as that would violate symmetry). In particular they cannot have differnt size: Let $\mathrm{fix}_G(E)$, $E\subseteq A$ finite, be a support for $f$. Find any $n<\omega$ such that $a_n^0, a_n^1\notin E$ and let $\pi$ be the permutation of $A$ that swaps $a_n^0, a_n^1$ and is the identity everywhere else (observe that $\pi\in\mathrm{fix}_G(E)$). Then $$f(a_n^1)=f(\pi(a_n^0))=\pi(f)(\pi(a_n^0))=\pi(f(a_n^0))$$ Now $\pi^+\upharpoonright f(a_n^0)$ is a bijection between $f(a_n^0)$ and $\pi(f(a_n^0))=f(a_n^1)$ in $\mathcal V$.
I am not quite sure at the moment how to turn this into a model of ZF with the same property...