axiom-of-choiceset-theory
We know that if $S$ is an amorphous set then $\mathcal{P}(S)$ is Dedekind-finite. I was wondering if the implication was reversible and decided to generalize a bit though I think the question in the title is perhaps a bit too open-ended.
If $\mathcal{P}(S)$ is an infinite Dedekind-finite set, is $S$ amorphous? Can we say much else about $S$?
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.
Well. There's not a whole lot we can say at this level of generality. For example, if $A$ is strongly amorphous, i.e. every partition of $A$ is almost entirely made of singletons, then it is easy to see that any permutation of $A$ can only move finitely many points, since it decomposes $A$ into orbits, and those must be finite.
On the other hand, if $A$ is a $2$-amorphous set, i.e. there is a partition into pairs, then there is also a permutation which simply switches the elements of each pair, and therefore has infinitely many nontrivial orbits.
On the third hand,1 if $A$ is amorphous, then any infinite partition of $A$ must also be amorphous, and so it is easy see that the behaviour across the different orbits of a given permutation must be the same almost everywhere.
Finally, and this is where it starts getting really weird, there's a lot of structure that is compatible with certain "level" of amorphous-ness2 sets. Given any first-order structure whose first-order definable subsets are finite or co-finite, and has sufficiently rich automorphism group (enough to witness strong homogeneity), we can arrange a model where this structure has an amorphous "version".
For example, an infinite dimensional vector space over a finite field. But also a set with some partitions that satisfy certain basic rules can be used.
So, if we have an infinite set with a partition into $3$-element sets, we can also add a function to this structure, which is a permutation that generates that partition; but we can also elect to not add such a function, and ensure that none exist, i.e. that a $3$-amorphous set will only have finitely supported permutations.
Therefore, at this "maximal" level of generality, of just talking about amorphous sets, there's not a whole lot that we can say. Sam Tarzi, a former student of Peter Cameron, wrote a Ph.D. thesis titled "Group Actions on Amorphous Sets and Reducts of Coloured Random Graphs" and they have a preprint (presumably derived from that work, although seems to have never been through the peer review process), titled "Group Actions On Amorphous Sets" that might be of interest to you.
Here's something that I couldn't find in the literature, but I took a few minutes to make the argument. But I am almost certain this should be known.
Theorem. If $A$ is amorphous, then $\operatorname{Sym}(A)$ is Dedekind-finite.
Proof. Suppose that $\{p_n\mid n<\omega\}$ is a sequence of distinct permutations of $A$. Recall that $[A]^{<\omega}$, the set of finite subsets of $A$, is Dedekind-finite and the each orbit of $p_n$ must be finite.
For a given $a\in A$, let $O(a)$ be the closure of $\{a\}$ under all the $p_n$s. We first claim that $O(a)$ must be finite. Since orbit of $a$ under each $p_n$ must be finite, there must be finitely many "types" of orbits, as $[A]^{<\omega}$ is Dedekind-finite; taking $O_0(a)$ as the union these orbits, we define $O_{n+1}(a)=\bigcup\{O(x)\mid x\in O_n(a)\}$, then each $O_{n+1}$ is finite, and this is an increasing sequence of sets in $[A]^{<\omega}$, so it must stabilise, and therefore $O(a)$ must be finite.
Next, as $\{O(a)\mid a\in A\}$ must form a partition of $A$, all but finitely have the same size. More importantly, though, all of the interesting properties that follow from how $p_n$, for whatever $n$, acts on $O(a)$ must also be the same for all but finitely many examples.
In particular, for a fixed $n$, all but finitely many $a$s have the same "type" of $p_n$ on $O(a)$. In other words, for a cofinite set of $a$ and $b$ there is a bijection mapping $O(a)$ to $O(b)$ which commutes with $p_n$.
So, letting $A_n$ be the [finite] set of "atypical $O(a)$", we have, by Dedekind-finiteness, that $\{A_n\mid n<\omega\}$ is just finitely many sets, repeating. Therefore $A^*=\bigcup A_n$ is a finite set, and for any $a,b\notin A^*$, $p_n$ has the same type on $O(a)$ and $O(b)$. But since each $O(a)$ is finite, there can only be finitely many of these types, so must be able to distinguish infinitely many $p_n$s just by examining $A^*$.
But now we ran into a bit of a problem! Since $A^*$ is finite, there can only be finitely many permutations of $A^*$, so we cannot distinguish infinitely many permutations at once!
So, the only way to resolve this is that there was no countably sequence of permutations, and therefore $\operatorname{Sym}(A)$ is Dedekind-finite as well. $\square$
Amorphous sets are weird, and apparently they have more than two hands. Don't get me started on the number of fingers each of those have.
Amorphity? amorphousity? amorphousnessicity? who knows...
Best Answer
There is one essential theorem here, due to Kuratowski.
It follows that $\mathcal P(X)$ being Dedekind-finite means that $X$ cannot be mapped onto $\omega$. From this it is easy to see that if $A$ and $B$ both cannot be mapped onto $\omega$, then $A\cup B$ cannot either. But if $A$ and $B$ are infinite, then $A\cup B$ is definitely not amorphous. This was pointed out by bof in the comments as well.
But now you might ask, perhaps it is true that every infinite set which cannot be mapped onto $\omega$ contains an amorphous subset? Maybe amorphous sets are the root of the problem.
The answer to that is also negative. It is possible that there is a linearly ordered space which is densely ordered, but has a Dedekind-finite power set. In particular, such set cannot have an amorphous subset. Since it's linearly ordered.
(See more here.)