This question is an attempt to chisel away at this earlier question of mine, which in retrospect may be rather intractable. Throughout, we work in $\mathsf{ZF}$.
Briefly (see the linked question for details), for a set of sentences $\Gamma$ and a filter $\mathscr{F}$, an infinite set $X$ is $\Gamma$-species reflecting to $\mathscr{F}$ iff the following are equivalent for each pair of sentences $\varphi,\psi\in\Gamma$:
$\vert M_X(\varphi)\vert\le \vert M_X(\psi)\vert$.
$\{n\in\omega: \vert M_n(\varphi)\vert\le \vert M_n(\psi)\vert\}\in\mathscr{F}$.
Here $M_A(\theta)$ is the set of isomorphism types of models of $\theta$ with underlying set $A$. In general, if $\Gamma$ is too broad then no $\Gamma$-species reflection (to any filter at all) is possible. However, for narrow $\Gamma$s, the situation is less clear. Specifically, let $\Gamma_{er}$ be the set of sentences using only binary relation symbols such that for each $\varphi\in\Gamma_{er}$ and each symbol $E$ occurring in $\varphi$ we have $\varphi\vdash$ "$E$ is an equivalence relation." The set $\Gamma_{er}$ is already rather interesting from the point of view of this question; for example, if $X$ is $\Gamma_{er}$-species reflecting to any filter then $X$ is not amorphous, and in fact this is already ensured by a two-element subset of $\Gamma_{er}$.
Is the statement "For every filter $\mathscr{F}$, there is an infinite $X$ which is $\Gamma_{er}$-species reflecting to $\mathscr{F}$" consistent with $\mathsf{ZF}$? What about $\mathsf{ZF+}$$\mathsf{BPI}$?
To be clear, I am expecting a negative answer here, so I've phrased the question strongly to make negative answers hopefully easier to prove. My interest in $\mathsf{BPI}$ is that species reflection to an ultrafilter yields very "linear" combinatorics, discussed in more detail at the above-linked question, so it seems likely that an inconsistency can be proved significantly more easily from $\mathsf{ZF+BPI}$ than from $\mathsf{ZF}$ alone.
