Let $v \in \mathbb{R}^3$ and $R$ be a rotation matrix (probably orthogonal is enough). I think the following property is true:
$$ [R\cdot v] = R \cdot [v] \cdot R^T$$ where $[v]$ is the skew symmetric form of the vector $v$. I should verify this by direct calculation, but I am not in the mood right now, hence I want to know if the following proof is true:
proof: let $u \in \mathbb{R}^3$ hence $$ [R\cdot v] \cdot \left( R \cdot u\right) = \left(R\cdot v\right) \times \left( R\cdot u\right) = R \cdot \left( v\times u\right) = R \cdot [v] \cdot R^T \cdot \left( R \cdot u\right)$$ therefore $$ [R\cdot v] \cdot \left( R \cdot u\right) = R \cdot [v] \cdot R^T \cdot \left( R \cdot u\right) $$ hence $\left([R\cdot v] - R \cdot [v] \cdot R^T \right) \cdot w = 0$ for all $w = R \cdot u$ with $u \in \mathbb{R}^3$. It follows that $\left([R\cdot v] - R \cdot [v] \cdot R^T \right) = O_{3 \times 3}$
Edit:
Suppose $v = \begin{bmatrix} a\\b\\c\end{bmatrix}$ then $[v] = \begin{bmatrix}0 &-c &b\\ c &0 &-a\\ -b &a &0 \end{bmatrix}$
