Diffeomorphism for mapping one SDE into another Let $Y_t,X_t$ be $(\Omega,\mathcal{F},\mathcal{F}_t,\mathbb{P})$-adapted Markov diffusion processes with valued in $\mathbb{R}^n$.  (When) does there exist a diffeomorphism $\phi:\mathbb{R}^n\to \mathbb{R}^n$ such that
$$
\phi(X_t)= Y_t \mathbb{P}-a.s?
$$
More generally, since the continuous image (in the above sense) of a diffusion process needs not be a diffusion process since the Ito-formula (and extensions thereof) fail.  What conditions do we need on $X,Y$ and $\phi$ such that $Y_t$ is just an $\mathbb{R}^n$-valued Markov process?
 A: This is more of a long comment. As far as I know, the first question is a very difficult (and interesting) one. Here are a few examples of diffusions in $\mathbb R^3$ that are not diffeomorphic in your sense, and may give an idea of the difficulties ahead:


*

*the standard Brownian motion in $\mathbb R^3$;

*the Brownian motion along planes $z=c$, i.e. $X_t=(B^1_t,B^2_t,c)$ for $c$ a constant and $B^1,B^2$ independent Brownian motions;

*the Brownian motion along planes with orthogonal drift, i.e. $X_t=(B^1_t,B^2_t,z_0+t)$;

*the Brownian motion along the (contact) distribution $dz-ydx=0$,¹ i.e.
$$X_t = \left(B^1_t,B^2_t,\int_0^tB^2_s\circ dB^1_s\right);$$

*the Brownian motion reflected along the plane $z=0$, i.e. $X_t = (B^1_t,B^2_t,Z_t)$ where $Z_t=B^3_t$ until $B^3$ reaches zero for the first time, then $Z_t=|B^3_t|$.


When one restricts to solution of SDEs of the form $dX_t=b(X_t)dt + \sigma(X_t)dB_t$ with $b,\sigma$ smooth and $\sigma$ of constant rank (points 1 to 4 above), we could imagine that the data of a distribution with a quadratic form on it, together with a vector field, would suffice to characterise the diffusion, but it isn't the case. For $\epsilon_1,\epsilon_2,\epsilon_3$ the standard basis of $\mathbb R^3$, $\epsilon_1 dB^1_t+\epsilon_2dB^2_t$ and
$$ \big( \epsilon_1\cos(x^1)+\epsilon_2\sin(x^1)\big)dB^1_t
 + \big(-\epsilon_1\sin(x^1)+\epsilon_2\cos(x^1)\big)dB^2_t$$
do not define identical diffusions, although the randomness propagates along the same planes $z=c$ and correspond to the same quadratic form. In short, the question you are asking


*

*contains strictly the question of whether two second degree operators are diffeomorphic, which

*contains strictly that of whether two Euclidean distributions are diffeomorphic, which

*contains strictly the (as far as I know) difficult topological question of whether two distributions are locally diffeomorphic,


so I think the first question is too broad as it stands.
Edit: Now that I think about it more, maybe my argument goes in fact in the other direction: given two unrelated diffusions $X$ and $Y$, there are so many different possibilities that it should be easy to show they are not diffeomorphic. In the examples 1 to 4 above, we can argue by considering iterated brackets, in the somewhat classical description of e.g. Hörmander.
Another example of this: if the diffusions are elliptic with the same quadratic part, for instance $X$ and $Y$ have generators $\frac12\Delta+b_X$ and $\frac12\Delta+b_Y$ for some smooth vector fields $b_X$ and $b_Y$, then a diffeomorphism sending $X$ to $Y$ would have to send the principal symbol of one generator to the other, and hence be an isometry. In this case it should be easy to see whether $b_Y$ is just a version of $b_X$ that got moved around.

¹ See the wikipedia page about contact structures for an illustration. Here I mean distribution in the sense “subbundle of the tangent bundle”.
