Consider surreal numbers as an H-field with operation of derivation.
In such setting for any surreal number $\alpha$ such that $0<\alpha<e^\omega$, $\partial(\alpha)<\alpha$ and for $\alpha>e^\omega$, $\partial (\alpha)>\alpha$.
Particularly, $\partial(\omega_1)$ is hugely greater than $\omega_1$.
I wonder, can we then define ordinals as $\omega_{n+1}=\partial(\omega_n)$, $n>0$?
Is this allowed by the automorphisms of the surreals?
It is not the case that $0<\alpha < \operatorname{e}^{\omega} \Longrightarrow \partial(\alpha)< \alpha$. For instance $\partial(\operatorname{e}^{\omega}-\omega^{-1})= \operatorname{e}^{\omega}+\omega^{-2}>\operatorname{e}^{\omega}-\omega^{-1}$.
However, it is the case that $0<\alpha < \mathbb{R}^{>0} \operatorname{e}^{\omega} \Longrightarrow \partial(\alpha)< \alpha$, and that $\mathbb{R}^{>0} \operatorname{e}^{\omega}<\alpha \Longrightarrow \partial(\alpha)> \alpha$. This holds indeed in any H-field (see Lemma 1.4 in the paper H-fields and their Liouville extensions by Matthias Aschenbrenner and Lou van den Dries).
It is also the case in all H-fields with small derivation that if $\alpha$ lies above the field of constants, then $\partial(\alpha) < \alpha^2$. So you can't go too far by derivating.
If by $\omega_1$ you mean the smallest uncountable ordinal, then it is a $\log$-atomic number, whence its derivative can be computed by the special formula for $\log$-atomic numbers. In that case, we have $\partial(\omega_1)=\exp(\log \omega_1 + \log \log \omega_1 +\log \log \log \omega_1 + \cdots)$, which is sometimes written as an infinite product $\omega_1 (\log \omega_1) (\log \log \omega_1) \cdots$. The same formula holds for all $\log$-atomic numbers $\lambda$ larger than all $\exp(\omega),\exp(\exp(\omega)),...$.
So $\partial(\omega_1)$ is indeed much larger than $\omega_1$ (and is not an ordinal), but infathomably smaller than $\omega_2$ in that every iterate of exp applied at $\partial(\omega_1)$ is still (much) smaller than $\omega_2$.
