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For every finite subset $A $ of $\mathbb C[[X]]$, does there exist a ring automorphism (bijective ring endomorphism ) $f$ of $\mathbb C[[X]]$ such that $f(a)=a, \forall a \in A$ but $f(\mathbb C) \ne \mathbb C$ ?

Related Are $\mathbb C$ , $\mathbb C[X]$ definable in $\mathbb C[[X]]$? because if $D \subseteq M$ is definable by a set $A \subseteq M$ then $f(D)=D, \forall f \in Aut_A M$ , and in the related case we mean definable by some finite subset of $M$.

UPDATE : The partial answer by Alex Kruckman here Given finite subset of $\mathbb C[[X]]$, does there exist a ring automorphism of $\mathbb C[[X]]$, fixing $A$, but not fixing $\mathbb C[X]$ set wise? shows that these two questions are equivalent.

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  • Related question: Is every algebraically closed subfield of $\mathbb{C}[[X]]$ contained in $\mathbb{C}$? – Alex Kruckman Mar 01 '18 at 19:25
  • @AlexKruckman: That's a good question and unfortunately I don't know the answer ... but why is it relevant ? – user Mar 01 '18 at 20:52
  • Because if $f$ is an automorphism of $\mathbb{C}[[X]]$ such that $f(\mathbb{C})\neq \mathbb{C}$, then either $f(\mathbb{C})$ or $f^{-1}(\mathbb{C})$ is an algebraically closed subfield which is not contained in $\mathbb{C}$. If no such subfield exists, no such automorphism exists. – Alex Kruckman Mar 01 '18 at 21:10

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