Is the following sentence equivalent to $\sf AC$ over the rest of axioms of $\sf ZF$?
For each infinite set $X$: if for all $y \in X$ we have $|y| < |X|$, then $| \bigcup X|\leq |X| $?
Note: The cardinality function $||$ here is defined after Scott's.
If not, then to which of the known choice principles this is equivalent to?
This comes in connection to this question.
Yes, this is like Tarski's result that if $\mathfrak{c}^2=\mathfrak{c}$ for every infinite cardinal $\mathfrak{c}$, then AC holds.
In fact, it suffices to suppose that $\mathfrak{c}\mathfrak{d}=\mathfrak{c}$ whenever $\mathfrak{c},\mathfrak{d}$ are cardinals and $\mathfrak{c}$ is infinite and $\mathfrak{d}<\mathfrak{c}$. (Note this follows from your hypothesis.)
For this, fix an infinite set $y$. It suffices to see that $y$ is wellorderable. We may assume that $y$ does not contain any ordinals. Let $\kappa$ be the least ordinal such that $\kappa\not\leq\mathrm{card}(y)$. Let $\mathfrak{c}=\mathrm{card}(y\bigcup\kappa)$. Then $\mathrm{card}(y)<\mathfrak{c}$, so by the hypothesis, $y\times(y\cup\kappa)$ has cardinality $y\cup\kappa$. So $y\times\kappa\leq y\cup\kappa$. But then the usual argument shows that $y$ is wellorderable, which suffices. That is, fix an injection $\pi:y\times\kappa\to y\cup\kappa$. Since $\kappa\not\leq\mathrm{card}(y)$, for each $x\in y$, there is some $\alpha<\kappa$ such that $\pi(x,\alpha)\notin y$, so $\pi(x,\alpha)\in\kappa$. Let $\alpha_x$ be the least such. Then define $f:y\to\kappa$ by setting $f(x)=\pi(x,\alpha_x)$, and note that $f$ is injective, so $y$ is wellorderable, as desired.
