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?



Diffeomorphism for mapping one SDE into another
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:


1. the standard Brownian motion in $\mathbb R^3$;
2. 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;
3. the Brownian motion along planes with orthogonal drift, i.e. $X\_t=(B^1\_t,B^2\_t,z\_0+t)$;
4. the Brownian motion along the (contact) distribution $dz-ydx=0$,¹ i.e.
$$X\_t = \left(B^1\_t,B^2\_t,\int\_0^tB^2\_s\circ dB^1\_s\right);$$
5. 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.




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¹ See the wikipedia page about [contact structures](https://en.wikipedia.org/wiki/Contact_geometry) for an illustration. Here I mean distribution in the sense “subbundle of the tangent bundle”.


