Is the following consistent with $\text{ZF}$? 


There exists a set $S=\{x\_1,x\_2,x\_3,...\}$ such that:


1. $|x\_{i+1}| < |x\_i|$
2. $\forall m,n \in S (|P(m)|=|P(n)|)$


Where cardinality $``||"$ is defined after Scott's.



Can we have an infinite sequence of decreasing cardinality all terms of which have equal sized power sets?
Yes.


For silly reasons.


Suppose that $X$ is a Dedekind-finite set, then $S(X)$, the set of all injective finite *sequences* from $X$ is also Dedekind-finite. Let $S\_n(X)$ denote the subset of $S(X)$ of sequences whose domain is *at least* $n$. It is easy to see why $S\_n(X)$ surjects onto $S(X)$. Simply erase the first $n$ coordinates.


This means, by arguments from my answer to your last question, that $S(X)\leq^\* S\_n(X)$ for all $n$. Since those are Dedekind-finite sets, and the inclusion is strict, they form the wanted sequence.


You can even be more clever than this, and for some chain in $\mathcal P(\omega)$ which has order type $\Bbb R$, define a sequence of order type $\Bbb R$ of Dedekind-finite sets, all of which have equipotent power sets.


