General problem: Let $X$ and $Y$ be processes taking values in $\mathbb{R}^n$ which solve the Stratonovich SDEs $$\partial X_t = \sigma(X_t) \partial W_t$$ $$\partial Y_t = \xi(Y_t) \partial B_t,$$ and further assume, for all $t\geq 0$, $X_t \stackrel{(d)}{=} Y_t$ and that $\sigma \sigma^T = \xi \xi^T$ identically (i.e. for all $x\in \mathbb{R}^n$).
Question: Do $X$ and $Y$ have the same Ito drift?
My effort: The Ito drift for $X$ is given by the formula $$ \mu^i = \frac12 \operatorname{Tr}(\sigma D[\sigma_{i,.}])$$ where $D[\sigma_{i,.}]$ is the Jacobian of the $i$-th row of $\sigma$. Note it can be seen that this can be written as $$2\mu = \nabla \cdot(\sigma \sigma^T) -\sigma \nabla \cdot(\sigma^T),$$ where the divergence operator is applied row-wise to matrices. The infinitesimal generator of $X$ is then $$\mathscr{G}_X f = \mu^T \nabla f + \frac12 \operatorname{Tr}(\sigma \sigma^T \nabla^2 f).$$
Let $a$ be the Ito drift term for $Y$. Then the two generators are equal if and only if for every smooth function $f:\mathbb{R}^n\to \mathbb{R}$, $$(\mu-a) \perp \nabla f$$ using the fact that $\sigma\sigma^T = \xi\xi^T$. This can be further simplified, again using the fact that $\sigma \sigma^T = \xi\xi^T$ to get the requirement that $$\mu - a = \frac12 \left(\xi \nabla \cdot(\xi^T)-\sigma \nabla\cdot (\sigma^T)\right) \perp \nabla f$$
So $\xi \nabla \cdot(\xi^T) = \sigma \nabla \cdot (\sigma^T)$ is a sufficient condition. Either of these conditions seem to be too much to ask to hold in general but I have not come up with a counterexample yet.
Update: If I am not mistaken, we can prove the following result in one dimension, as I outlined in the comments below:
If $\partial X = \sigma(X)\partial W$ and $\partial Y = \xi(Y)\partial B$ are Stratonovich processes with smooth diffusion coefficients $\sigma, \xi\in C^1(\mathbb{R})$ such that $\sigma(x)^2 =\xi(x)^2$ for all $x\in \mathbb{R}$, then they have equal Ito drifts, and thus if $X_0=Y_0$ a.s., then $X_t \stackrel{(d)}{=} Y_t$ for all $t\geq 0$.
Proof: Since $\sigma(x)^2=\xi(x)^2$ identically, we have $2\sigma(x) \sigma'(x) = 2\xi(x)\xi'(x)$ identically. But the Ito drift of $X$ is $\mu_X = \frac12 \sigma(x)\sigma'(x)$ and the Ito drift of $Y$ is $\mu_Y = \frac12 \xi(x) \xi'(x)$, thus it follows that $\mu_X=\mu_Y$, identically. So these processes have the same infinitesimal generator $$\mathscr{G} f = \frac12 \sigma(x)\sigma'(x) \frac{\partial f}{\partial x} + \frac12 \sigma(x)^2 \frac{\partial^2 f}{\partial x^2}.$$ Thus if $X_0=Y_0$ almost surely, it follows by Feynman-Kac that for any Borel set $A\subset \mathbb{R}$, that $\mathbb{P}(X_T \in A|X_0) = \mathbb{P}(Y_T\in A|Y_0)$, for any $T>0$.