# $\sum_{i=1}^nx_i\frac{\partial}{\partial x_i}$ is invariant under all matrices

Consider the differential operator $$D:=\sum_{i=1}^nx_i\frac{\partial}{\partial x_i}$$. Then I claim that $$D(u\circ A)(x)=D(u)(Ax)$$. The proof is as follows: First note that $$Du(x)=\langle x, \nabla_x u(x)\rangle$$ with the gradient $$\nabla_x$$. By the chain rule we have that $$\nabla_x (u\circ A)(x)=A^T\nabla_yu(y)|_{y=Ax}$$. Thus $$\langle x,\nabla_x(u\circ A)(x)\rangle=\langle x,A^T\nabla_yu(y)|_{y=Ax}\rangle=\langle Ax,\nabla_yu(y)|_{y=Ax}\rangle=\langle y,\nabla_yu(y)\rangle|_{y=Ax}$$ proving that $$D(u\circ A)(x)=D(u)(Ax)$$. Is there a mistake in the proof? I would rather expect only invariance under orthogonal transformations. Thanks in advance.