The question I am going to ask is really to satisfy my curiosity, as I am not at all an expert of the subject and do not plan to really work on it. Hence, if you think the question is not suitable for MO, I'll delete it. I did some browsing but could not locate any answer, but maybe I missed something simple.

Take ZF as a basis. If $Z\subset\omega$, denote by $AC(Z)$ the axiom of choice for sets of (finite) sets whose cardinality is in $Z$:

$AC(Z)$: If $x\not=\emptyset$ and for all $y\in x$ we have $|y|\in Z$, then there is a choice function on $x$.

Thus $AC(\omega)$ is the axiom of choice for sets of finite sets, and $AC(\{n\})$ (for $n\in\omega$) the axiom of choice for sets of $n$-elements sets. The implications between $AC(Z)$ for finite $Z$ and $AC(\{n\})$ have apparently been well studied in the first years of the 70s (and before by Tarski and Mostowski), I don't know to which point the question is entirely solved.

Just to give an idea, some (classical ?) implications I found in the literature are:

\begin{align*}AC(\{2\})&\Leftrightarrow AC(\{4\}),\\ AC(\{3,7\})&\Rightarrow AC(\{9\}),\\ AC(\{2,3,7\})&\Rightarrow AC(\{14\}). \end{align*}

It is easy to see that $AC(\{n\cdot m\})\Rightarrow AC(\{n\})$: Given $x$ containing $n$-element sets, there is a choice function on $\{y\times m\,:\,y\in x\}$ by $AC(\{n\cdot m\})$, the projection on the first factor gives a choice function on $x$. Thus, if $Z$ contains all the multiples of some given $n$, then $AC(Z)\Leftrightarrow AC(\omega)$.

My question is whether there is some kind of converse:

If $AC(Z) \Leftrightarrow AC(\omega)$, can we say something about $Z$ ?

For instance, I am sure that $Z$ cannot be finite, but can it be Dedekind finite ? Must it be "big" in any appropriate sense ?