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In Automorphisms of $\mathbb C_p$, K. Conrad showed that there exist uncountably many $\mathbb Q_p$-automorphisms of $\mathbb C_p$. I have a quite similar question.

Let $(a_n)_{n\in\mathbb N}$ and $(b_n)_{n\in\mathbb N}$ be sequences in $\mathbb Q_p$. Does there exist an uncountable set $S$ of continuous automorphisms of $\mathbb C_p$ that leave invariant $\{b_n\mid n\in\mathbb N\}$ and such that the set of images of the sequence $(a_n)_{n\in\mathbb N}$ by the elements of $S$ is uncountable. I assume that for all $n\in\mathbb N$, $b_n\notin\mathbb Q(\smash{a_j}_{\mid j\in\mathbb N})$.

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  • $\begingroup$ I do not think this is a "quite similar question" to your previous one. Here you want $S$ to contain continuous automorphisms of $\mathbf C_p$. If you mean they are continuous using the ordinary topology on $\mathbf C_p$ then they fix all elements of $\mathbf Q_p$ (they fix all elements of $\mathbf Q$ by algebra and then all of $\mathbf Q_p$ by continuity) so the image of each $a_n$ by $S$ is just $a_n$. If you did not want elements of $S$ to be continuous for the ordinary topology on $\mathbf C_p$ then what topology did you have in mind? $\endgroup$
    – KConrad
    Mar 13, 2019 at 8:12
  • $\begingroup$ By the way the cardinal of the automorphism group of $\mathbf{C}_p$ as $\mathbf{Q}_p$-algebra is not only uncountable, but $2^{2^{\aleph_0}}$. Its subgroup of continuous automorphisms has a natural topology of infinite second countable profinite group, and has cardinal $2^{\aleph_0}$. $\endgroup$
    – YCor
    Mar 13, 2019 at 9:51
  • $\begingroup$ By the way, the word is 'automorphism' (ending in 'ism'), not 'automorphim' (ending in 'im'). I have edited to correct. $\endgroup$
    – LSpice
    Mar 13, 2019 at 17:02
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    $\begingroup$ The continuous automorphism group is $\text{Gal}(\overline{\mathbf{Q}_p}/\mathbf{Q}_p)$ the easy part is $T = \lim_{n \to \infty} \mathbf{Q}_p(\zeta_{p^{n!}-1}, p^{1/(p^{n!}-1)})$ the maximal tamely ramified extension whose Galois group is $\prod_{\ell \ne p} \text{Aff}(\mathbf{Z}_{\ell})$ and the complicated part is the remaining tower of totally ramified extensions $\endgroup$
    – reuns
    Mar 13, 2019 at 19:30

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