# Ordinal enumeration in ordered Mostowski model - does it not need the global choice?

In Jech's axiom of choice he proves following lemma (lemma 4.5(b) in his book):

There is an injective mapping from $M$ to $\mathrm{Ord}\times \operatorname{fin}(A)$, where $A$ is a set of atoms and $M$ is the ordered Mostowski model.

I have a doubt to his proof. Its proof considers the orbit $$\operatorname{orb}(x) = \{\pi x : \pi\in G\}$$ (for symmetric group $G$) and argues that we have an ordinal enumeration of class of all orbits, since $\operatorname{sym}(\operatorname{orb}(x)) = G$. However I fail to get why it holds. In the book we starts from the model of ZFA with choice and from this we can deduce

A set $x$ is well-ordered in $M$ iff $\operatorname{fix} x$ is in the normal filter.

If the above assertion holds for classes $x$, we can proceed the proof of the lemma. However, I can not find any reason that holds for classes, unless the global choice holds (so every class can be well-ordered.) Is there a detour to prove the lemma without global choice? I would appreciate your answer, thanks!

It's almost necessary to get s linear order for the entire model. But if you just want to order each set separately, you can find some large enough ordinal where the proof works for $x$ using that ordinal.
But if, for example, your original $V$ cannot even be linearly ordered due to its kernel being "complicated enough", then there is no way that $M$ will be linearly ordered either.