Can I rotate divergence away? / Can get divergence from a rotation? Let $v(x,y)$ be a two-dimensional vector field and let $R(x,y, \theta)$ be the two-dimensional rotation matrix which rotations a vector field around $(x,y)$ an angle $\theta$.
The following two questions are probably very much related:
Question 1 : Does a $v$ exists such that $\nabla \cdot v = 0$, but $\nabla \cdot (R v) \neq 0$ for some $x,y,\theta$?
Question 2 : Does a $v$ exists such that $\nabla \cdot v \neq 0$, but $\nabla \cdot (R v) = 0$ for some $x,y,\theta$?
Thank you for any insight in advance.
My intuition tells me that the answer is probably simply "No" to both questions, but I can't prove it.

Illustrating the example by Shuhao Cao: 
$$v(x,y) = (x,y) ;\qquad u(x,y) = (-y,x)$$
Here $\operatorname{div} v\equiv 2$ and $\operatorname{div} u\equiv 0$. The two fields are related by rotation by $90$ degrees: red has divergence, blue doesn't. 

