How many automorphisms are there for $ \langle \omega , < \rangle $?

Question

I'm not sure how to start this, although I expect there to be an upper bound of $2^{\aleph_0}$. ($\aleph_0^{\aleph_0} = 2^{\aleph_0}$)

Update:

Could the answer just be $1$? Since the ordinals are well ordered, it should just be rigid, right?

What's your current lower bound? Which is to say, how many automorphisms do you know of? — Arthur, Nov 15 '22 at 12:54
1 (identity function), and now that I think about it, it could be the only automorphism as ordinals are well ordered sets — amlearn369, Nov 15 '22 at 13:11
Try to prove it’s the only one. Hint: given an automorphism $f$, can it be $f(0)\neq 0$? After that, what about $f(1)\neq 1$?… — Lorenzo Pompili, Nov 15 '22 at 13:18
Once you miss something in the range, there's no going back to pick it up later. — JonathanZ, Nov 15 '22 at 13:30

score 3 · Accepted Answer · answered Nov 15 '22 at 14:18

Suppose that $f$ is an automorphism of $\omega$ which is not the identity map. Since $\omega$ is well-ordered, there is a least $n$ such that $f(n) \neq n$. By the choice of $n$ we have $f(m) = m$ for all $m < n$. Since $f$ is one-to-one, $f(n) > n$. Since $f$ is onto, there is a $k > n$ such that $f(k) = n$. But then $n < k$ but $f(k) < f(n)$. The function $f$ is not an automorphism.

How many automorphisms are there for $ \langle \omega , < \rangle $?

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