The partial preorder on $\mathbb N$ generated by the finite axioms of choice Let $\mathsf C_n$ denotes the statement: 
for any family $\mathcal F$ of $n$-element sets there exists a choice function (i.e., a function $f:\mathcal F\to\bigcup\mathcal F$ such that $f(F)\in F$ for all $F\in\mathcal F$).
It is known that $\mathsf C_2\Rightarrow \mathsf C_4$ in ZF. 
This fact suggests introducing a partial preorder $\preceq$ on the set $\mathbb N$ of positive integers defined by $n\preceq m$ if $\mathsf C_m\Rightarrow \mathsf C_n$ in ZF. 
Also we can write that $n\cong m$ if $\mathsf C_n\Leftrightarrow \mathsf C_m$.
It is easy to show that $n\preceq m$ if $n$ divides $m$. So, $1\preceq n$ for any $n\in\mathbb N$ and $2\preceq n$ for any even number $n$.
On the other hand, $\mathsf C_2\Rightarrow \mathsf C_4$ implies that $2\cong 4$.
What else is known about the partial preorder $\preceq$? Maybe there exists a precise (arithmetic) description of this preorder.
A  more specific question: is $2^n\cong 2$ for any $n\in\mathbb N$?
I know that similar questions were studied by Mostowski, Tarski, Truss, Jech so maybe the answer is already known?
 A: In Jech's "The Axiom of Choice", at the end of Chapter 7, he formulates the following condition on two natural numbers $n>m$:

(S) There is no decomposition of $n$ into $p_1+\ldots+p_s=n$ such that $p_i>m$ is a prime number for all $i$.

And he goes on to prove that if $\mathsf{C}_k$ holds for all $k\leq m$, then (S) implies $\mathsf{C}_n$, and moreover if (S) fails, then there is a model of $\mathsf{C}_k$ for all $k\leq m$, but $\lnot\mathsf{C}_n$.

So, for example, if the Goldbach conjecture is true, then $\mathsf{C}_2$ implies nothing more than $\mathsf{C}_4$, since in that case every even number other than $2$ is the sum of two primes, and other than $4$ these primes have to be odd.
So this is now a question about number theory, rather than set theory, and I will let the experts in that field make their remarks.
Emil Jeřábek notes in the comments that every $n\geq 8$ is the sum of some amount of $3$s and $5$, both prime and both odd, so $\mathsf{C}_2$ does not extend its reach beyond $\mathsf{C}_4$.
