Recall that a set is amorphous iff it is infinite but has no partition into two infinite subsets. I'm interested in the possible structure (in the sense of model theory) which an amorphous set can carry. For example, if $T$ is a complete first-order theory with an amorphous model then $T$ must be strongly minimal; however, the converse fails badly (consider $Th(\mathbb{C};+,\times)$).
I'm curious whether second-order logic does a better job. For $T$ a complete second-order theory with no finite models, say that $T$ is second-order strongly minimal iff every second-order-definable subset of every model $\mathcal{A}\models T$ is either finite or cofinite in $\mathcal{A}$. For example, $Th(\mathbb{C};+,\times)$ is not second-order strongly minimal since $\mathbb{Z}\subset\mathbb{C}$ is second-order definable.
EDIT: by "complete" I mean "maximally satisfiable:" a second-order theory $T$ is complete iff it is satisfiable and for each second-order sentence $\varphi$, either $\varphi\in T$ or $\neg\varphi\in T$.
To be reasonably concrete, my main question is the following:
Question 1: Is it consistent with $\mathsf{ZF}$ that every second-order strongly minimal theory has an amorphous model?
(More broadly, I'm interested in whether there is a reasonably natural logic $\mathcal{L}$ such that consistently with $\mathsf{ZF}$ every "$\mathcal{L}$-strongly-minimal" theory has an amorphous model. Second-order logic just seems like a good candidate at the moment.)
There's also a "virtual" version of the question which makes sense even in the presence of choice:
Question 2: Is it consistent with $\mathsf{ZFC}$ that for every second-order strongly minimal theory $T$, there is some (set) symmetric extension in which $T$ has an amorphous model?
The exact relationship between these two questions isn't clear to me. That said, an affirmative answer to question 2 seems much more plausible than an affirmative answer to question 1.