Let $k$ be a field and $(A,m)$ be the completion of the local ring of a smooth point of a $k$-variety. Let $x\_1,x\_2\in m\backslash m^2$ be regular elements. I am interested in knowing if one can find a $k$-linear automorphism of $A$ which takes the ideal $(x\_1)$ to $(x\_2)$. 


If $k$ is perfect, it is easy to see that this can be done, since there is a $k$-linear embedding $A/m\hookrightarrow A$ which induces a $k$-linear isomorphism 
$$ A \cong (A/m)[|x\_1,...,x\_n|] $$ for any choice of regular sequence $(x\_1,..,x\_n)$ in $A$. 


When $k$ is not perfect, I can hardly find any non-trivial $k$-automorphism of $(A,m)$.



Automorphisms of complete local rings
Here is a counterexample. Fix $a\in k\smallsetminus k^p$ and take the completion of the affine plane at the point $(0,a^{1/p})$. In other words, $A$ is the completion of the local ring $k[u,v]\_{(u,v^p-a)}$.


Now take $x\_1=u$ and $x\_2=v^p-a$. Then $A/(x\_1)$ is the completion of $k[v]$ at $(v^p-a)$, while $A/(x\_2)\cong k(a^{1/p})[[u]]$. The residue field $k(a^{1/p})$ lifts to the second quotient, but not to the first, as you can check. Hence $A/(x\_1)\not\cong A/(x\_2)$ as $k$-algebras.


