Let $B_1, B_2$ be two independants Brownian loops on $R^3$.
Does it almost surely exists a continuous isomorphism $\phi$ of $\mathbb{R}^3$ onto itself such that $\phi(Im(B_1))=Im(B_2)$? Can we chose $\phi$ homotopic to identity?
(just remark that this seems trivially true in $\mathbb{R}^n$ for $n>3$, since then the map $t\mapsto B_t$ is bijective, extends to a small tubular neighborhood, and then to the full $\mathbb{R}^3$ since there is "no knot" on dimension $4$. Thus every brownian loop is homotopic to a simple circle, thus they are homotopic between them. As opposed to this, in dimension $3$, a Brownian loop knots and intersects itself a lot.)