Obtaining the conjugated momentum operator after coordinate transformation I am mostly uncertain if my definition of the momentum operator and the resulting commutator is correct after a simple coordinate transformation. Lets say we have in our first coordinate system the following definitions,
$$
H = -\frac{\hbar^2}{2m}\frac{d^2 }{dx^2} + V(x),\\
p_x =  -i\hbar \frac{d}{dx}
$$
Now we introduce the following coordinate $y=m^{1/2}x$.
My first question is do I obtain the correct momentum operator by simply using the coordinate transformation laws? I.e. is the new momentum given by the follwing?
$$
p_y = -i\hbar \frac{dy}{dx}\frac{d}{dy} 
$$
which yields in for this particular case
$$
p_y = -i\hbar m^{1/2}\frac{d}{dy}.
$$
Using this momentum operator and the new coordinate in the commutator yields
$$[y,p_y] = i\hbar m^{1/2} $$
which looks weird. Should I have defined the new momentum operator instead as $-i\hbar \frac{d}{dy}$ instead to obtain the common commutation relation?
It also occurred to me that I do not know how to obtain the conjugated momentum operator after a general coordinate transformation, while I have no trouble to do so for the classical quantities using the Lagrangian and $p_y=\frac{\partial L(x(y), \dot x(y))}{\partial \dot y}$, yet $\hat p_y = ?$
Is there a simple prescription to obtain the new momentum operator based on an given  invertible coordinate transformation?
 A: What OP is really asking is whether the canonical momentum of a given coordinate  $\hat{y}$ is of the form
$$\hat{p}_y=-i\hbar\frac{\partial}{\partial y}\tag{1}$$
The answer is yes. Making a canonical coordinate transformation leaves Poisson Brackets invariant. The transformation you proposed is indeed canonical$^1$.
Regarding the factor the factor $m^{\frac{1}{2}}$ in your commutator, the problem comes from assuming your new momentum is the old one expressed in the new variables, that causes problems due to the reasons I explained above. Given a position operator, whether it is angular or any other kind of position variable, the canonical conjugate momentum is defined as in $(1)$, or equivalently through the canonical commutation relations
$$[\hat{y},\hat{p}_y]=i\hbar$$
that in turn is equivalent to say that it is the infinitesimal generator of translations.
I'll make an example where something similar happens to make things more clear. If you call $\varphi$ the angle on the $xy$ plane, the $z$-component of angular momentum is the conjugate momentum
$$\hat{L}_z=-i\hbar\frac{\partial}{\partial\varphi}\implies [\hat{\varphi},\hat{L}_z]=i\hbar$$

$^1$ In this particular case, this is a point transformation.
A: The comments and the preliminary answer from User Feynman_00 helped me to see where I went wrong. My error was assuming that the new momentum operator is the same as old one expressed in the new coordinates.
$$\hat p_y \neq \hat p_x=-i\hbar \frac{d}{dx}=-i\hbar \frac{dy}{dx}\frac{d}{dy} = -i\hbar m^{1/2}\frac{d}{dy}=\hat p_x(y)$$
I carried the error over to the commutator by
$$
i\hbar = [x,p_x] \neq [y, \hat p_x(y)]=i\hbar m^{1/2}
$$
If the classical Lagrangian has the form
$$L_x = \frac{1}{2}m\dot x^2 - V(x)\\
L_y = \frac{1}{2}\dot y^2  - \tilde V(y)
$$
we can derive
$$
p_y = \dot y \\
p_x = m\dot x  = m^{1/2} \dot y = m^{1/2}p_y
$$
Assuming that this carries over to the quantum operators, yields the proper result
$$\begin{aligned}
\hat p_y &= m^{-1/2} \hat p_x \\
&= m^{-1/2} \hat p_x(y) = -i\hbar \frac{d}{dy} \Rightarrow \\
[\hat x, \hat p_x] &= [\hat y, \hat p_y] = i\hbar
\end{aligned}$$
