Finite bit flips cannot defeat a free ultrafilter

No finite set $F$ belongs to $U$. Otherwise, since $U$ is not principal, no singleton $\{n\}$ belongs to $U$; F2 then puts every $\omega \setminus \{n\}$ in $U$. Intersecting these complements for the finitely many $n\in F$ puts $\omega \setminus F$ in $U$, so propriety is contradicted by $F\cap (\omega \setminus F)=\mathrm{\varnothing }$. This includes $F=\mathrm{\varnothing }$, which is excluded directly by propriety. By F2, every cofinite set therefore belongs to $U$.

Put $D=\omega \setminus (X\mathrm{△}Y)$. By F3 this is exactly the set on which $X$ and $Y$ agree, and it is cofinite by the Given hypothesis; hence $D\in U$ by step 1.1. If $X\in U$, then $X\cap D\in U$, while agreement gives $X\cap D\subseteq Y$, so upward closure gives $Y\in U$. Exchanging $X$ and $Y$ proves the reverse implication. Thus the displayed equivalence holds, including $X=Y$, $X=\mathrm{\varnothing }$, and $X=\omega$.

A flip of only finitely many bits replaces $X$ by some $Y$ with finite $X\mathrm{△}Y$, so step 2.1 preserves its membership status in $U$ and supplies no contradiction. Feferman's automorphism instead sends the selected real to a finite modification of $\omega \setminus X$: invariance then transfers its membership to the complement, which conflicts with F2. The distinction is between a finite flip and an infinite tail flip, not between two descriptions of the same automorphism.