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 $[q,p]=1$; indeed, 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  $-[A,B]y^2$.

**********************************************************************

**Motivation:** Please see [this][1] question, in order to understand the motivation for my above question.


Any hints and comments are welcome!
(I have also asked the above question in [MSE][2], but have not received any comments yet). 


  [1]: https://mathoverflow.net/questions/334897/a-non-commutative-analog-of-a-result-concerning-a-jacobian-pair
  [2]: https://math.stackexchange.com/questions/3281015/is-a-specific-endomorphism-of-a-1-an-automorphism