I vaguely recall the following result, but can't find it or prove it on my own; I'd like to check if it, or something similar to it, is actually true. My memory is that the proof is rather hard, FWIW.
Vague memory: it is consistent with $\mathsf{ZF}$ that there are linear orders $X,Y$ such that
every cofinal subset of $X$ is in (non-order-preserving) bijection with some cofinal subset of $Y$ and every cofinal subset of $Y$ is in (non-order-preserving) bijection with some cofinal subset of $X$, but
there is no linear order $Z$ which cofinally embeds into both $X$ and $Y$.
Basically, in the absence of choice the class of (Scott) cardinalities of cofinal subsets of a linear order can fail to determine the class of ordertypes of cofinal subsets of that linear order, in an extremely strong way: two linear orders can have identical "coarse cofinality behaviors" but completely disjoint "fine cofinality behaviors." One immediate observation is that neither $X$ and $Y$ can have a well-ordered cofinal subset, but beyond that not much is clear to me.