Action of infinite symmetric groups on iterated power sets Let $X$ be an infinite set, and $k \ge 1$ be a natural number.  We work without the axiom of choice.
Let $G_0$ be the full symmetric group on $X$, and let $G_1$ be the
full symmetric group on ${\cal P}(X)$, the power set of $X$. Both
these groups act on ${\cal P}^k(X)$ (the $k$-fold power set $\mathcal{P}(\mathcal{P}(\cdots(X)\cdots)$ of $X$)
in a natural way, with $G_0$ a subgroup of $G_1$. So in ${\cal P}^k(X)$ the orbits of $G_1$ split into orbits of $G_0$.  

Is the following true in ZF: for all $k\ge 1$, for all distinct elements $a\neq b$ of ${\cal P}^k(X)$
  that belong to the same $G_1$-orbit, there exists $\sigma \in G_1$ such that
  $\sigma(a)$ and $\sigma(b)$ belong to distinct $G_0$-orbits?

A positive result would have nice consequences for the model theory of Russell-Ramsey typed set theory, but I'll say nothing about that for the moment!
 A: Here's a comment (no longer an answer) about the case $k=2$, assuming the axiom of choice, and in particular assuming that $X$ is presented in one-on-one correspondence with some cardinal.  Perhaps this can be generalized substantially.
Now $a,b\in{\cal P}^2(X)$ are in the same $G_1$ orbit iff they have the same cardinality.  [Update: their complements should also have the same cardinality, and the construction below assumes infinite complements.]  So for this case, we want to know:  Given $a,b\subset {\cal P}(X)$ of the same infinite cardinality, are there sets $c,d\subset {\cal P}(X)$ of that same cardinality but in different $G_0$ orbits?
The answer to that question is yes. For $S\in{\cal P}(X)$, let
\begin{align}
f(S)&=\{s+4:s\in S\}\cup\{1,2,3\}\\
c&=\{f(S):S\in a\}\cup \{\{1,2\},\{2,3\},\{3,1\}\}\\
d&=\{f(S):S\in b\}\cup \{\{1\},\{2\},\{3\}\}
\end{align}
Clearly $a,b,c,d$ are all of the same cardinality, and:


*

*i) $\exists x\in X$ which appears in all but one element of $c$.

*ii) $\nexists x\in X$ which appears in all but one element of $d$.


These properties are invariant under $G_0$ actions, so $c$ and $d$ are in different $G_0$ orbits.
