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-ness^{2} 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

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.