This question is addressed here. In particular, the diffusion that satisfies the required symmetry property is a special case of a $\nu$-symmetric diffusion. By itself, this symmetry property does not uniquely specify the diffusion. For example, the requisite property holds for any $d$-dimensional diffusion of the form $$ d X_t = \operatorname{div}M(X_t) dt + \sqrt{2} B(X_t) dW_t $$ where $B(x)$ is a $d \times d$ matrix and $M(x) = B(x) B(x)^T$. Finally, when $\nu$ is additionally normalizable, a related question was asked here.

This question is addressed here. In particular, a diffusion whose transition densities are symmetric is a special case of a $\nu$-symmetric diffusion, and by itself, this symmetry does not uniquely specify the diffusion. For example, the symmetry property holds for any $d$-dimensional diffusion of the form $$ d X_t = \operatorname{div}M(X_t) dt + \sqrt{2} B(X_t) dW_t $$ where $B(x)$ is a $d \times d$ matrix, $M(x) = B(x) B(x)^T$ for all $x \in \mathbb{R}^d$, and $(\operatorname{div}M)_i = \sum_{j=1}^n \partial M_{ij}/ \partial x_j$ is the divergence of a second-order tensor field. To specify the diffusion, it suffices to specify the geometry of the noise, because then the corresponding drift field is uniquely determined. When the density $\nu$ is additionally normalizable i.e. $\int_{\mathbb{R}^d} \nu(x) dx < \infty$, then a related question came up here.