Choosing elements of linear orders

Is it consistent with ZF that there can be a countable family of linear orders, each isomorphic to $\mathbb Z$ (that is, every element has a unique predecessor and successor, and any two elements have finitely many elements between them), such that one cannot choose an element from each of them?

(Random musing inspired by this question.)

Yes, it is consistent. Let me sketch an argument using atoms, then the Jech-Sochor transfer theorem can do its magic (since it's a bounded statement).

Start with a countable set of atoms which is the disjoint union of $A_n$'s each countably infinite. Fix $\leq_n$ on $A_n$ which is a linear order of type $\omega^*+\omega$ (or $\Bbb Z$, in other words).

The automorphism group is $\mathscr G$, of permutations of the atoms which preserve both the partition into the $A_n$'s and their linear orders. In other words, we take a sequence of automorphisms of $\leq_n$ (which are finite shifts), and they each act on the relevant $A_n$. Let us denote a permutation by $\pi$ and $\pi_n$ is the permutation on the $n$-th coordinate.

We define a filter of subgroups in the following way. Given $H$ a subgroup of $\mathscr G$, $H$ is in the filter if and only if there exists a finite $E\subseteq\omega$, such that whenever $\pi\in\mathscr G$ for which $\forall n\in E$, $\pi_n=\operatorname{id}$, then $\pi\in H$. (For example taking $E=\varnothing$ we have that $\{\pi\in\mathscr G\mid\forall n\in\varnothing(\pi_n=\operatorname{id})\}=\mathscr G$ itself.)

Some easy propositions:

1. Each $A_n$ is symmetric, and each $\leq_n$ is symmetric.
2. The enumerated partition $\{A_n\mid n\in\omega\}$ is symmetric, and the enumeration of the orders is also symmetric.

And the important point: There is no choice function from the $A_n$'s.

Sketch for a proof: If $f$ was such a choice function, it would have a finite support $E$ as above, now taking any $k\notin E$ we can apply any $\pi$ which fixes $f$ but $\pi_k\neq\operatorname{id}$ to get a contradiction.

(This is the usual argument.)

Now we can use the Jech-Sochor theorem. Or we could have used forcing to begin with, it would just make things harder to write down properly in this case.