The presentation of Cohen's first model that I'm most familiar with is to start with a forcing extension by $Add(\omega,\omega)$, consider the group of automorphisms of $Add(\omega,\omega)$ that permute the first $\omega$ and the filter of subgroups that fix finitely many elements of $\omega$, and define the model of hereditarily symmetric names to be those names fixed by some group of automorphisms in the filter (hereditarily).
I know that the set $A=\{a_i\}_{i\in\omega}$ of generic reals added is Dedekind-finite, since for any name of an injective function $f$ into $A$ there must have some $i$ and $j$ outside the subgroup fixing $f$ so that $a_i$ and $a_j$ can be permuted and we get the contradiction that if $f(n)=a_i$ then also $f(n)=a_j$.
My question is whether the same contradiction applies to any linear ordering of $A$. Since $A$ is a set of reals, it can be linearly ordered in the Cohen model. If we consider the set of ordered pairs $l=\{<a_i,a_j>|a_i<_\mathbb{R}a_j\}$, it seems like we could also find some $i$ and $j$ outside the subgroup fixing $l$ and get that both $a_i<a_j$ and $a_j<a_i$ must be true. ZF should guarantee that $l$ is in the model, but this argument seems to guarantee that it isn't. Any help resolving this confusion would be appreciated!
No. The reason is, of course, the linear ordering of the reals is generally taken to be the lexicographic ordering (i.e., the first $n$ for which the two reals differ decides which one is "bigger").
At the level of the maximum condition we don't have any information as to which reals lies where. And when we do apply a permutation to a condition which, say, decided that $a_0$ will be after $a_3$, that bit of information "moves" with the names of the reals. So, if $p\Vdash\dot a_3<\dot a_0$, say because $p(0,0)=1$ and $p(3,0)=0$, and we take $\pi$ to be the permutation which switches between $0$ and $3$, then $\pi p(0,0)=0$ and $\pi p(3,0)=1$, so $\pi p\Vdash\dot a_0<\dot a_3$.
In particular, the "canonical" name for the lexicographic order is going to be stable under any permutations for that reason.