Let ACE (Axiom of Choice for Equinumerous sets) be the following choice principal:
If $S$ is a set of non-empty sets such for any $X,Y\in S$ there is a bijection from $X$ to $Y$, then $S$ has a choice function.
I want to know if ACE implies AC or is strictly weaker.
Motivation
In the case of choice from families of finite sets, the analog of ACE is strictly weaker than AC, i.e. there is a model where for each $n$, every set of $n$-element sets has a choice function, but there is a countable collection of finite sets (of unbounded size) with no choice function. You can get this by taking a permutation model with a set of $p$ atoms for each prime $p$, being acted on by the group $\prod_{p\text{ prime}} \mathbb{Z}/p\mathbb{Z}$, with finite supports.
However, ACE implies every set of finite sets has a choice function. To see this, suppose $S$ is a set of finite non-empty sets. Then $\{X\times\omega | X\in S\}$ is a collection of countable sets, so has a choice function by ACE. It's easy to build a choice function for $S$ from this.
Best Answer
For any set $A$ and nonempty subset $B \subseteq A$, $B \times A^\omega$ is equinumerous with $A^\omega$: there is an injection from $B \times A^\omega$ to $A^\omega$ because $B \subseteq A$, and there is an injection from $A^\omega$ to $B \times A^\omega$ because $B$ is not empty, so by the Schröder-Bernstein theorem there is a bijection. By ACE, this implies there is a choice function on $\{B \times A^\omega | B \subseteq A\}$ (excluding the empty set), which implies there is a choice function on the power set of A (excluding the empty set), which is equivalent to the axiom of choice. So ACE implies AC.