Is this true in the ordered Mostowski model?

Question

The ordered Mostowski model $M$ is the permutation model used by Mostowski to show that the axiom of choice is independent from the ordering principle. The set $A$ of atoms of this model is isomorphic to the dense linear ordering $\mathbb{Q}$, the group $\mathcal{G}$ is taken to be the automorphsim group of $A$, and the normal ideal is taken to be the set of all finite subsets of $A$. For $x\in M$ and $B\subseteq A$, $B$ is said to be a support of $x$ if every $\pi\in\mathcal{G}$ fixing $B$ pointwise also fixes $x$. Is the following true for this model?

For every $x\in M$ and $\pi\in\mathcal{G}$, if $\pi(x)=x$, then $\mathrm{fix}(\pi)$ is a support of $x$, where $\mathrm{fix}(\pi)$ is the fixed points of $\pi$.

Answer 1

If I remember correctly, in the same paper that introduced the ordered model $M$, Mostowski showed that, if two finite sets of atoms support the same element $x\in M$, then so does their intersection. It follows that every $x\in M$ has a unique smallest support, say $S(x)$. Thanks to the uniqueness, any automorphism $\pi$ that fixes $x$ must also fix $S(x)$, and in fact must fix $S(x)$ pointwise because of the linear ordering of the atoms. So fix$(\pi)$ is a superset of $S(x)$ and therefore supports $x$.