Models of ZFA corresponding exactly with a particular class of groups I recently read [1], in which Blass exhibits a correspondence between:


*

*Permutation models of ZFA in which the axiom of choice (AC) fails but the Boolean prime ideal theorem (BPIT) holds; and

*Nontrivial extremely amenable topological groups with small open subgroups.


This correspondence is demonstrated in two theorems (Theorems 1 and 2 below) with the intermediate step of Ramsey filters. But the correspondence is not exact, in a sense I'll explain.

Background
A Ramsey filter on a group $G$ is a filter $\mathcal{F}$ of subgroups of $G$ containing a subgroup $H$ for which each $K \le H$ lying in $\mathcal{F}$ is a Ramsey subgroup of $H$, i.e. for any $2$-colouring $c : H/K \to 2$ and any finite subset $A \subseteq H/K$ there is some $h \in H$ for which $h \cdot A$ is monochromatic (where the action of $H$ on $H/K$ is the natural one).
A topological group has small open subgroups if every neighbourhood of the identity contains an open subgroup. It is extremely amenable if it is Hausdorff and whenever it acts continuously on a compact Hausdorff space, the action has a fixed point.
Theorem 1. If $G$ is a topological group with small open subgroups and $\mathcal{F}$ is the filter of all open subgroups of $G$, then $G$ is nontrivial and extremely amenable if and only if every subgroup in $\mathcal{F}$ is a Ramsey subgroup of $G$ and $\{ 1 \} \not \in \mathcal{F}$
Theorem 2. If $M$ is a model of ZFA induced by a group $G$ and a normal filter $\mathcal{F}$ of subgroups of $G$, in which all atoms are symmetric and every subgroup in $\mathcal{F}$ stabilizes some set $x \in M$, then the Boolean prime ideal theorem holds in $M$ if and only if $\mathcal{F}$ is a Ramsey filter of subgroups of $G$; and moreover the axiom of choice fails in $M$ if and only if $\{ 1 \} \not \in \mathcal{F}$.

Observation
Every subgroup in $\mathcal{F}$ being a Ramsey subgroup of $G$ is a stronger condition on $\mathcal{F}$ than $\mathcal{F}$ simply being a Ramsey filter: it's what you get if you stick $G$ in the place of $H$ in the definition of a Ramsey filter I gave above. Blass observed this fact, and demonstrated that extreme amenability cannot correspond with the Ramseyness of the filter of open subgroups.
But another question can be asked. In some sense the correspondence goes one way: any nontrivial extremely amenable topological groups with small open subgroups gives rise to a model of ZFA+BPIT+(¬AC); but we don't (automatically) have the opposite direction. Instead of weakening the condition on $\mathcal{F}$ in Theorem 1, I'd like to know what happens if we (try to) strengthen the condition on $\mathcal{F}$ in Theorem 2 and thus obtain a (more) exact correspondence.

Question
Is it known if there is some statement $\phi$ strictly intermediate in strength between AC and BPIT which forces, for a model of ZFA+$\phi$+(¬AC) generated by $G$ and $\mathcal{F}$ satisfying the hypothesis of Theorem 2, that $\mathcal{F}$ satisfy the hypothesis of Theorem 1 rather than simply being a Ramsey filter?
A slight weakening of the question: is there a statement $\phi$ as above such that, for any model of ZFA+$\phi$+(¬AC) generated by $G$ and $\mathcal{F}$ satisfying the hypotheses of Theorem 2, there exist $G'$ and $\mathcal{F}'$ which generate the same model and $\mathcal{F}'$ satisfies the hypothesis of Theorem 1.

I hope this isn't a silly question. I've given it some thought but haven't made much progress. Also, I typed this out in quite a hurry, so please forgive any mistakes I might have made.
[1] Andreas Blass, Partitions and Permutation Groups, Contemporary Mathematics (2011), no. 558, pp. 453–466.
 A: The answer to the first question is negative, because the permutation model generated by $G$ and $\mathcal F$ doesn't really "see" the group $G$.  Consider the following "shrinking" operation.  Replace $G$ by a subgroup $H$ that belongs to $\mathcal F$ and replace $\mathcal F$ by its restriction to $H$ (i.e., $\{K\in\mathcal F:K\subseteq H\}$).  Then the permutation model doesn't change, but whether $\mathcal F$ satisfies the hypotheses of Theorem 1 can change.  Specifically, if $\mathcal F$ is a Ramsey filter, as witnessed by a subgroup $H$, then using this $H$ in the shrinking construction produces the same permutation model but now with a group and filter satisfying the hypotheses of Theorem 1.  So no property of the permutation model can exactly match the hypotheses of Theorem 1.  (You could think of "Ramsey filter" as the hypothesis of Theorem 1 made invariant under (inverse) shrinking so that it can be a property of the permutation model.)
The same observations give a positive answer to your second question, with $\phi$ being BPIT itself.  The desired $G'$ and $\mathcal F'$ are obtained by suitable shrinking as above.
