Recently I asked a question about whether a second-order analogue of strong minimality could correspond to amorphous satisfiability (= having a model whose underlying set cannot be partitioned into two infinite pieces). I missed a fairly obvious fact: roughly speaking, that in a second-order way we can assert the existence of a certain type of structure on the domain prohibiting amorphousness (e.g. a linear order or a non-surjective endofunction) without making any such structure definable.
I'd like to rephrase my original question to avoid this. One idea which seems promising is to demand something like a second-order witness property: whenever our theory asserts the existence of some structure on the domain there should be some definable such structure. The existence of a second-order sentence saying "The domain is amorphous" causes a bit of difficulty in making the question nontrivial, however (e.g. any maximally satisfiable second-order theory either proves that the domain is amorphous, in which case it obviously is amorphously satisfiable, or proves that there is a partition of the domain into two infinite piecs at which point the second-order witness property kills any type of minimality).
The easiest way to get around this, in my opinion, is to restrict attention to monadic theories. My question is whether it is consistent with $\mathsf{ZF}$ that every satisfiable monadic second-order theory $T$ with the following properties is amorphously satisfiable:
$T$ is negation-complete: for each $\varphi$, either $\varphi\in T$ or $\neg\varphi\in T$.
$T$ is "second-order strongly minimal:" no model of $T$ has a definable-with-(element-)parameters bi-infinite subset.
$T$ has the "second-order witness property:" whenever $\exists X\varphi\in T$, there is some unary predicate symbol $U$ such that $\varphi[X/U]\in T$.
Note that something like the witness property is necessary here, per Harry West's answer to my original question.