I thought it is straight forward to do, but turned out to be very difficult. Normal cross product properties do not work here.
It is commonly used in 3D Problems involving multiple rotated frames, but couldn't find a source explicitly proving it. Tried random values in Matlab and the equation holds numerically.
as response to Rahul's comment, do you mean so? :
Changing notation, we have to prove: $$ \left[\mathbf{R}\mathbf{k}\right]_\times = \mathbf{R}[\mathbf{k}]_\times \mathbf{R}^\text{T}. $$ Multiply both sides by an arbitrary vector $\mathbf{v}\in\mathbb{R}^3$, $$ \left[\mathbf{R}\mathbf{k}\right]_\times\mathbf{v} = \mathbf{R}[\mathbf{k}]_\times \mathbf{R}^\text{T}\mathbf{v}. $$ Replace the skew-symmetric matrices with actual cross products, $$ \left(\mathbf{R}\mathbf{k}\right)\times\mathbf{v} = \mathbf{R}(\mathbf{k} \times \mathbf{R}^\text{T}\mathbf{v}), $$ and distribute (this is possible for rotation matrices) $$ \left(\mathbf{R}\mathbf{k}\right)\times\mathbf{v} = (\mathbf{R}\mathbf{k}) \times (\mathbf{R}\mathbf{R}^\text{T}\mathbf{v}) $$ For a rotation matrix, we know $\mathbf{R}^\text{T}=\mathbf{R}^{-1}$ (equivalently $\mathbf{RR}^\text{T}=\mathbf{I}$), so $$ \left(\mathbf{R}\mathbf{k}\right)\times\mathbf{v} = (\mathbf{R}\mathbf{k}) \times \mathbf{v}, $$ so we've shown the two sides are the same for all $\mathbf{v}\in\mathbb{R}^3$.