If a set $S$ can be partitioned into $\mathfrak{n}$ amorphous sets, can $S$ be partitioned into $\mathfrak{m}$ infinite sets? Assume $\sf{ZF}$. A set $A$ is amorphous if it is infinite and not the disjoint union of two infinite sets. The existence of an amorphous set is inconsistent with $\sf{ZFC}$ (even $\sf{ZF}+\sf{AC}_{\omega}$).
The statement $P_{F}(\mathfrak{n},\mathfrak{m})$ where $\mathfrak{n}$ and $\mathfrak{m}$ are cardinals is defined as follows:

Let $A_{i},i\in I$ be a collection of pairwise disjoint amorphous sets where $|I|=\mathfrak{n}$, and let $A=\bigcup\limits_{i\in I}A_{i}$. Then there exist nonempty pairwise disjoint sets $B_{j},j\in J$ with $|J|=\mathfrak{m}$ such that $A=\bigcup\limits_{j\in J}B_{j}$.

Similarly, the statement $P_{I}(\mathfrak{n},\mathfrak{m})$ is defined as

Let $A_{i},i\in I$ be a collection of pairwise disjoint amorphous sets where $|I|=\mathfrak{n}$, and let $A=\bigcup\limits_{i\in I}A_{i}$. Then there exist pairwise disjoint infinite sets $B_{j},j\in J$ with $|J|=\mathfrak{m}$ such that $A=\bigcup\limits_{j\in J}B_{j}$.

Lastly, the statement $P_{A}(\mathfrak{n},\mathfrak{m})$ is defined as

Let $A_{i},i\in I$ be a collection of pairwise disjoint amorphous sets where $|I|=\mathfrak{n}$, and let $A=\bigcup\limits_{i\in I}A_{i}$. Then there exist pairwise disjoint amorphous sets $B_{j},j\in J$ with $|J|=\mathfrak{m}$ such that $A=\bigcup\limits_{j\in J}B_{j}$.

For which $\mathfrak{n},\mathfrak{m}$ are $P_{F}(\mathfrak{n},\mathfrak{m})$, $P_{I}(\mathfrak{n},\mathfrak{m})$, and $P_{A}(\mathfrak{n},\mathfrak{m})$ true, false, and undecidable in $\sf{ZF}$?
 A: As a very partial answer, note that things are quite simple in the finite-index-set case for $P_A$: no nontrivial instances of the principle can hold.
Suppose $$X=\bigsqcup_{i\in I}A_i=\bigsqcup_{j\in J}B_j$$ where $A_i, B_j$ are all amorphous. Consider the function $$f: I\rightarrow J$$ sending each $i$ to the unique $j\in J$ such that $A_i\cap B_j$ is infinite. This is indeed well-defined: $A_i$ is amorphous, so we can't have $A_i\cap B_j$ and $A_i\cap B_{j'}$ both be infinite for $j\not=j'$. The function $f$ is injective, since each $B_j$ is amorphous: if $f(i)=f(i')=j$ for $i\not=i'$, then $B_j\cap A_{i}$ and $B_j\cap A_{i'}$ would be two disjoint infinite subsets of $B_j$.
By symmetry, we similarly get an injection $J\rightarrow I$, hence a bijection $I\cong J$ by Cantor-Bernstein (which does not require choice).

More generally, "finite" is really a proxy for "does not admit a finite-to-one embedding from an amorphous set," which in turn is the same as "does not have an amorphous subset." So actually we have:

If $I\not\cong J$ and neither $I$ nor $J$ have an amorphous subset, then no set is both an $I$-fold and a $J$-fold disjoint union of amorphous sets.

