Tarski: the Axiom of Choice is equivalent to the statement that $A\times A\approx A$ for every infinite set $A$, so extending Hessenberg's theorem from the alephs to arbitrary sets is exactly as strong as choice

If $A$ is infinite then every $n\in \omega$ injects into $A$: by induction along [L8] on the statement "there exists an injection $n\to A$", the empty function serving at $n=0$, and an injection $f:n\to A$ never being surjective, since $A\approx n$ would make $A$ finite, so that some $a\in A\setminus f[n]$ exists and $f\cup \{(n,a)\}$ injects $n\cup \{n\}$ into $A$; the statement carried through the induction is an existence statement, so no family of injections is selected.

Assume (1) and let $A$ be infinite; then $A$ is well-orderable by [L3], $\kappa =\mid A\mid$ is a cardinal with $A\approx \kappa$ by [L4], and $\kappa$ is infinite, since $\kappa \in \omega$ would make $A$ finite; so $A\times A\approx \kappa \times \kappa \approx \kappa \approx A$ by [L5], [L1] and [L4], which is (2).

Assume (2) from here on, let $A$ be infinite and put $\kappa =\mathrm{\aleph }(A)$; then $\kappa$ is a cardinal by [L2], and $\omega \le \kappa$, because $\kappa \in \omega$ would make $\kappa$ a natural number, which injects into $A$ by step 1.1 and contradicts [L2]; so $\kappa$ is an infinite cardinal.

The set $B=A\bigsqcup \kappa$ is infinite: $\xi \mapsto {\xi }^{\mathrm{\prime }}$ injects $\kappa$, hence also $\omega \subseteq \kappa$, into $B$ by step 2.1, so $B\approx n$ for some $n\in \omega$ would inject $n\cup \{n\}\subseteq \omega$ into $n$, which [L7] forbids; therefore (2) applies to $B$ and we may fix a bijection $f:B\times B\to B$.

Then $A$ is well-orderable. Exactly one of two situations holds. If some $a\in A$ has $f(a^{\mathrm{\prime }},{\xi }^{\mathrm{\prime }})\in \{b^{\mathrm{\prime }}:b\in A\}$ for every $\xi \in \kappa$, then sending $\xi$ to the unique $b\in A$ with $f(a^{\mathrm{\prime }},{\xi }^{\mathrm{\prime }})=b^{\mathrm{\prime }}$ is an injection $\kappa \to A$, which [L2] forbids. Otherwise every $a\in A$ admits some $\xi \in \kappa$ with $f(a^{\mathrm{\prime }},{\xi }^{\mathrm{\prime }})\in \{{\eta }^{\mathrm{\prime }}:\eta \in \kappa \}$; let ${\xi }_a$ be the $\in$-least such $\xi$, which is determined and not chosen by [L6], and let ${\eta }_a\in \kappa$ be given by $f(a^{\mathrm{\prime }},{\xi }_a^{\mathrm{\prime }})={\eta }_a^{\mathrm{\prime }}$. The map $a\mapsto ({\xi }_a,{\eta }_a)$ is then an injection $A\to \kappa \times \kappa$, since $({\xi }_a,{\eta }_a)=({\xi }_b,{\eta }_b)$ gives $f(a^{\mathrm{\prime }},{\xi }_a^{\mathrm{\prime }})=f(b^{\mathrm{\prime }},{\xi }_b^{\mathrm{\prime }})$ and hence $a^{\mathrm{\prime }}=b^{\mathrm{\prime }}$ by injectivity of $f$; composing with a bijection $\kappa \times \kappa \to \kappa$ from step 2.1 and [L1] injects $A$ into $\kappa$, and transporting the ordinal well-order of $\kappa$ back along that injection well-orders $A$, a nonempty subset of $A$ receiving the preimage of the $\in$-least element of its image.

A finite $A$ is well-orderable outright, being equinumerous with a natural number by [L7], so under (2) every set can be well ordered by step 4.1, and (1) follows by [L3]; with step 1.2 the two statements are equivalent over ZF.