I have been struggling to prove or find a counterexample to the following:
Fix a field $k$ of characteristic $0$. Let $R:= k[x_1,...,x_n]$ and $A$ be the ring of differential operators on $R$ (so $A$ is isomorphic to the $n$th Weyl algebra; in particular, it is just a ring of polynomials in the operators $\hat{x_1}, ..., \hat{x_n}, \partial_1,...,\partial_n$, where $\hat{x_i}$ is the multiplication by $x_i$ operator and $\partial_i$ is the ith partial derivative operator). We say an operator $D \in A$ is locally nilpotent if for all $f \in k[x_1,...,x_n]$ there exists a positive integer $n$ so that $D^n(f) = 0$. Let $\phi: A \to A$ be an automorphism of $A$. If $D$ is a locally nilpotent $k$-derivation in $A$, is the same true of $\phi(D)$? If this is true, does the statement hold in general when $R$ is a finitely generated integral domain over $k$?
Any thoughts would be appreciated.
No, this is not true - in fact, automorphisms of $A$ need not even send derivations to derivations. Let $n=1$ so that the Weyl algebra is $k\langle x,\partial\rangle/(\partial x- x\partial=1)$. Then $\partial$ is locally nilpotent: if $p$ is a polynomial of degree $n$, $\partial^{n+1}p=0$, while $x$ acts injectively on $k[x]$. But there is an automorphism sending $\partial\mapsto x$ and $x\mapsto -\partial$.