= Solution
Assume the <axiom of choice>. By the <well-ordering theorem>, every set is bijective with an <ordinal>, and hence with an <initial ordinal>. Any two ordinals are comparable, so for any sets $K,L$ either $K$ injects into $L$ or $L$ injects into $K$. Transitivity is immediate, and antisymmetry on cardinalities follows from the <Cantor-Schröder-Bernstein theorem>. Thus cardinal comparison is a total order.
Conversely, assume the <cardinal comparability principle>. Given any set $X$, let $h(X)$ be its Hartogs ordinal. By <Hartogs theorem>, there is no injection $h(X)\to X$. Comparability therefore supplies an injection $X\to h(X)$. Pulling the ordinal order back to $X$ gives a well-order of $X$. Hence every set can be well-ordered, and the <well-ordering theorem> implies the axiom of choice.
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