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According to this question, the size of $S_{\mathbb N}$, the set of all bijections on $\mathbb N$, is at least $2^{\aleph_0}$. So $|S_{\mathbb N}| \ge 2^{\aleph_0}$

For each bijection $f:\mathbb N\rightarrow \mathbb N$, define $\le_f$ as $x \le_f y$ iff $f(x) \le f(y)$. This relation is clearly reflexive, antisymmetric, and total. It also seems to be transitive, since $[x \le_f y]\land [y \le_f z]\Leftrightarrow [f(x) \le f(y)]\land [f(y) \le f(z)]\Rightarrow f(x) \le f(z) \Leftrightarrow x \le_f z$. Finally, for any set $A\subseteq\mathbb N$, the least element of $A$ by $\le_f$ is the least element of $f^{-1}(\min f[A])$, which always exists because the standard ordering on $\mathbb N$ is a well-order. So for each bijection $f:\mathbb N\rightarrow \mathbb N$, there is a well ordering $\le_f$, and thus $S_{\mathbb N}$ injects into $\omega_1$. So we have $|\omega_1|\ge|S_{\mathbb N}|$. Ordering of cardinals is transitive, so $|\omega_1| \ge 2^{\aleph_0}$.

Now, obviously there must be a break in logic here, because $|\omega_1|\ge 2^{\aleph_0}$ is going to need the continuum hypothesis somewhere. The transitivity and well-ordering parts are most suspect, and maybe I'm missing something obvious, but I can't seem to come up with a counterexample. So where exactly did I go wrong?

eyeballfrog
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    The map $\le_\bullet$ sending $f\mapsto \le_f$ is a map from $S_{\Bbb N}$ to the set of well-orderings on $\Bbb N$ (which has in fact cardinality $2^{\aleph_0}$). However, the map sending $f$ to the type of order of $\le_f$, which would be the map $S_{\Bbb N}\to\omega_1$ you wish for, is not injective and it's in fact constant equal to $\omega$. – Sassatelli Giulio May 31 '22 at 14:14
  • Probably just a typo: the relation $\le_f$ is not symmetric, but antisymmetric. – jjagmath May 31 '22 at 14:20
  • @SassatelliGiulio Oh goodness, you're right. I guess that's the answer then. – eyeballfrog May 31 '22 at 14:22

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