Let $k$ be a field of characteristic zero, and let $A_1(k)$ be the first Weyl algebra, namely, the associative non-commutative $k$-algebra generated by $x$ and $y$ subject to the relation $yx-xy=1$.
Suppose that the following map $f$ is a $k$-algebra endomorphism of $A_1$: $(x,y) \mapsto (f(x):=p,f(y):=q)$, where $p=Ay$ and $q=x+By$, $A,B \in A_1(k)$, $y$ does not divide $A$.
As a $k$-algebra endomorphism of $A_1(k)$, we have $[q,p]=1$; just apply $f$ to $yx-xy=1$.
Is it true that $f$ is actually an automorphism of $A_1(k)$? In particular, is it true that necessarily $p=- y$?
My partial answer: $1=[q,p]=[x+By,Ay]=[x,Ay]+[By,Ay]=-[Ay,x]+[By,Ay]$ $=-(A[y,x]+[A,x]y)+[By,Ay]=-(A+[A,x]y)+[By,Ay]=-A-[A,x]y+[By,Ay]$
Denote: $E:=[By,Ay]$. Then, $E=[By,Ay]=B[y,Ay]+[B,Ay]y=-B[Ay,y]-[Ay,B]y=$ $-B(A[y,y]+[A,y]y)-[Ay,B]y=-B[A,y]y-[Ay,B]y=$ $-B[A,y]y-(A[y,B]y+[A,B]y^2)=$ $-B[A,y]y-A[y,B]y-[A,B]y^2$.
So we have $E=-B[A,y]y-A[y,B]y-[A,B]y^2$.
Then, $1=-A-[A,x]y-B[A,y]y-A[y,B]y-[A,B]y^2$.
Write $A=a_ny^n+\cdots+a_1y+a_0$, $a_j \in k[x]$, $a_0 \neq 0$ (since we have assumed that $y$ does not divide $A$).
We see that $a_0=-1$.
Now, the highest $(0,1)$-term of $-A-[A,x]y-B[A,y]y-A[y,B]y-[A,B]y^2$should be zero, and by considerations of $(0,1)$-degrees, it equals the $(0,1)$-highest term of $-[A,B]y^2$.
Remark: Please see this question (or this same question), in order to understand the motivation for my above question. I have also asked this question in MO.
Any hints and comments are welcome!