In ZF, can a Dedekind-finite set (or even an amorphous set) be smaller than a non-trivial partition of its elements? For context: A set is amorphous if every subset of it is either finite or cofinite but not both (in particular is infinite). A set is D(edekind)-finite if every injection into itself is a surjection. One can easily that $X$ is D-finite iff $\omega \preceq X$, an thus, that an amorphous set must be an infinite D-finite set. (Finite here means in bijection with some $n < \omega$).
Question 1: Is it possible to have a D-finite set $X$ and nontrivial (i.e different from $\{\{x\} : x \in X\}$) partition $\Pi$ such that $X \preceq \Pi$? (Clearly $X$ can't be finite)
Of course, in presence of the axiom of choice this is not possible, since if we have
an injective function $f : X \to \Pi$, then, if $g : \Pi \to X$ is a choice function we get
that $g \circ f$ is an injective function, and thus since $X$ is D-finite, a surjection also. Concluding that $\Pi$ must be trivial. We also conclude of this that $\Pi$ must have infinite non-singleton elements.
Question 2: Can $X$ be amorphous?
One can prove that for an amorphous set $A$ and a partition $\Pi$ of $A$ into finite sets there must exist a unique $n \in \omega$ such that all but finitely many elements in $\Pi$ are of size $n(\Pi)$. $A$ is called bounded amorphous if $n(\Pi)$ is bounded over the possible partitions, otherwise is called unbounded. It is called strictly amorphous if $n(\Pi) = 1$ for all such partitions. See for example Wikipedia's entry, or Truss's paper "The structure of amorphous sets" for more on this.
Question 3: Can $X$ be bounded amorphous?
I've proved that $X$ can't be strictly amorphous, but not much more.
Edit: I think the answer to 1 is a "yes": I've realized that the existence of such partition (assuming X D-finite) is equivalent to the existence of a surjective $f : X \to X$ which is not injective. And I think is known that there are D-finite sets with non-injective surjections into itself.
 A: Yes, no, and consequently also no.
The first one is actually a theorem: if there is an infinite Dedekind-finite set, then there is one which can be mapped onto a strictly larger set. That means that the surjection defines the partition you are looking for.
Suppose that $A$ is Dedekind-finite, then $S_{inj}(A)$, which is the set of all injective finite sequences from $A$, is also Dedekind-finite. For if it wasn't, then we would have a countably infinite set of enumerated finite sets, and their union is therefore a countably infinite subset of $A$, and that is impossible.
Now take, for example $S_{inj}(A)$ without the empty sequence. Then the function erasing the first coordinate from each non-empty sequence is surjective onto $S_{inj}(A)$. Indeed, taking any subset which contains "all sequences of length $n\in I$" for some fixed infinite $I\subseteq\omega$ will work in a similar way.
For the second question, note that if $A$ is amorphous and $f\colon A\to X$ is surjective, then $X$ must be amorphous (or finite, if you'd like to require that amorphous implies infinite). For otherwise, simply partition $X$ into two infinite subsets and look at their preimages, both would have to be infinite subsets of $A$ which are disjoint.
How does that help us? Well, if $X$ is infinite, then the preimage of each point is finite. Now, starting with some $x\in X\setminus A$, consider $f^{-1}(x)=A_0$, then $f^{-1}(A_0)=A_1$ is also finite, and it must be distinct from $A_0$, and then we can continue by recursion and define a countably family of finite subsets of $A$. But this is impossible as the power set of an amorphous set is itself Dedekind-finite.
Note, however, that if $X$ is amorphous, its partitions are either finite; almost all singletons; or incomparable with $X$ itself. So we can get a partition that is incomparable, just not a partition that is strictly bigger.
