Let $A\rightarrow R$ be a local homomorphism of Noetherian strict henselian local rings with completions $\hat{A},\hat{R}$.
Let $u\in R^\times, x\in R$ be such that there is a unique $\hat{A}$-linear automorphism of $\hat{R}$ sending $x\mapsto ux$. Must there exist an $A$-linear automorphism of $R$ sending $x\mapsto ux$?
Context: I want to say that if $A[x]\rightarrow R$ is the strict henselization of $A[x]$ at $x = 0$, and $g\in\text{Aut}_A(R)$ has order $n$ invertible in $A$, then there is an automorphism ("reparametrization") $\alpha\in\text{Aut}_A(R)$ such that $g(\alpha(x)) = \zeta_n\alpha(x)$ for some primitive $n$th root of unity. I can show the existence of this $\alpha$ only in the completion $\hat{R}$, where it is given by $x\mapsto ux$ for some $u\in R^\times$. More precisely, I can find such a $u\in R^\times$ such that $g(ux) = \zeta_n ux$, and the explicit form of the completion shows that this defines an automorphism, but I don't know if it comes from an automorphism of $R$.
