What is the relation between the infinitesimal generator $\mathcal{A}$ of an Ito diffusion and the transition density function of the process?
More specifically, let's consider this subcase. Suppose I have the transition density $p_t(x|y)$ of a process $X_t$ with infinitesimal generator $\mathcal{A}$. Let $\kappa\in\mathbb{R}$ be fixed. What is the transition density of the process generated by $\kappa\mathcal{A}$? Is other information needed?
My thoughts: from PDEs, scaling the heat equation essentially scales time in the fundamental solution. So I think we should get something like $\tilde{p}_{\kappa t}(x|y)$ for the density of the new process. Not sure how to show it though.
By definition, the (strong) generator of a time-homogeneous (or "autonomous" if you like) Markov process is
$$\mathcal{A}[u](x)=\left. \frac{d}{dt} [E[u(X_t) \mid X_0=x] \right |_{t=0}.$$
So
$$\kappa \mathcal{A}[u](x) = \kappa \left. \frac{d}{dt} E[u(X_t) \mid X_0=x] \right |_{t=0} \\ = \left. \frac{d}{d(t/\kappa)} E[u(X_t) \mid X_0=x] \right |_{t=0} \\ = \left. \frac{d}{dt} E[u(X_{\kappa t}) \mid X_0=x] \right |_{t=0}. $$
That is, $\kappa \mathcal{A}$ is just the generator of $X_{\kappa t}$. So its transition density is just $p_{\kappa t}$. In effect you have, as you said, just accelerated time by a factor of $\kappa$.
Notably, doing so multiplies $\mu$ by $\kappa$ and multiplies $\sigma^2$, not $\sigma$, by $\kappa$. That is, multiplying through by $\kappa$ in the generator does not do the same thing as multiplying through by $\kappa$ in the SDE, because $\kappa dB_t$ is not the same as $dB_{\kappa t}$.