Is seven-dimensional cross product rotationally invariant?

Question

For three-dimensional cross product, the following property holds true: \begin{equation} (R\mathbf x) \times (R \mathbf y)=R(\mathbf x \times \mathbf y) \end{equation} where $R\in SO(3)$.

Is the analogous property (with $R\in SO(7)$) true for seven-dimensional cross product?

Answer 1

Counterexample: $$\mathbf{x}:=(1,0,...,0)^T,\mathbf{y}:=(0,...,0,1)^T\implies \mathbf z:=\mathbf{x}\times\mathbf{y}=(0,0,0,0,0,1,0)^T\\ R_{ij}:={1\over\sqrt{2}}(\delta_{i1}\delta_{j1}+\delta_{i2}\delta_{j2}-\delta_{i1}\delta_{j2}+\delta_{i2}\delta_{j1})+\delta_{ij}(1-\delta_{i1})(1-\delta_{i2})\\ R\mathbf{x}={1\over\sqrt{2}}(1,1,0,0,0,0,0)^T\\ R\mathbf{y}={1\over\sqrt{2}}(-1,1,0,0,0,0,0)^T\\ (R\mathbf{x})\times(R\mathbf{y})=(0,0,1,0,0,0,0)^T\\ R\mathbf z=\mathbf z $$