Show that if $A^*A=I$ then $\Delta(u\circ A)=\Delta u\circ A$ 
Let $u\in C^2(\Bbb R^n,\Bbb R)$ and $A\in\mathcal L(\Bbb R^n)$. Show that if $A^*A=I$ then $\Delta(u\circ A)=\Delta u\circ A$.

Here $\Delta$ is the laplacian operator (here is a similar question but this doesnt help me so much).
What I did was:
$$\partial^2(u\circ A)=(\partial^2 u\circ A)[\partial A]^2+(\partial u\circ A)\partial^2 A$$
and because $\partial^2 A=0$ and $\partial Ax=A$ we found that
$$\partial^2(u\circ A)(x)=(\partial^2 u\circ A)(x)A^2$$
However $A^2$ is not necessarily the identity, so I dont know exactly how to continue. Some help will be appreciated, thank you.
 A: $\newcommand{\dx}{d{\bf x}}$Here's a neat way to do this using integration by parts:
Note that $\Delta(u\circ A)-(\Delta u)\circ A$ is continuous, and so we have $\Delta(u\circ A)=(\Delta u)\circ A$ if for all $\psi\in\mathscr{C}^1_c(\mathbb{R}^n)$ we have
$$\int_{\mathbb{R}^n}\Delta(u\circ A)\psi~d{\bf x}=\int_{\mathbb{R}^N}((\Delta u)\circ A)\psi~d{\bf x}.$$
So fix some $\psi\in\mathscr{C}^1_c(\mathbb{R}^n)$, and let $\Omega=B({\bf 0},r)$ be a large enough ball containing the support of $\psi$. Note that $\Omega$ is $O(n)$-invariant. Now we just compute using integration by parts
\begin{align}
\int_{\mathbb{R}^n}\Delta(u\circ A)\psi~\dx&=\int_{\Omega}\Delta(u\circ A)\psi~\dx \\
& = -\int_{\Omega}\nabla(u\circ A)\cdot\nabla\psi~\dx \tag{integration by parts}\\
& = -\int_{\Omega}(A^{*}(\nabla u))\cdot\nabla\psi~\dx \tag{chain rule} \\
& = -\int_{\mathbb{R}^n}(A^{*}(\nabla u))\cdot\nabla\psi~\dx.
\end{align}
Similarly, we compute
\begin{align}
\int_{\mathbb{R}^n}((\Delta u)\circ A)\psi~\dx & = \int_{\Omega}((\Delta u)\circ A)\psi~\dx \\
& = \int_{\Omega}(\Delta u)(\psi\circ A^{-1})~\dx \tag{change of variables} \\
& = -\int_{\Omega}\nabla u\cdot\nabla(\psi\circ A^{-1})~\dx \tag{integration by parts} \\
& = -\int_{\Omega}\nabla u\cdot((A^{-1})^{*}\nabla\psi)~\dx \tag{chain rule}\\
& = -\int_{\Omega}\nabla u\cdot(A\nabla\psi)~\dx \tag{$A$ is orthogonal} \\
& = -\int_{\mathbb{R}^n}\nabla u\cdot(A\nabla\psi)~\dx
\end{align}
Finally, we just need to show that for any vectors $v,w\in\mathbb{R}^n$ that $Av\cdot w=v\cdot A^*w$. This is straightforward to see though using the matrix notation of the dot product
$$ Av\cdot w = (Av)^{*}w=v^{*}A^{*}w=v\cdot A^*w.$$
Concluding, we have shown that both of the above computations are equal for any $\psi\in\mathscr{C}^1_c(\mathbb{R}^n)$, and so we deduce the desired result of $\Delta(u\circ A)=(\Delta u)\circ A$ in $\mathbb{R}^n$.
A: I finally solved it using properties of the orthogonality of $A$ and expressing $\partial^2$ through the inner product and the hessian. I will add the answer just for the record.
Unfortunately I cant accept the answer of @yousuf because I dont know something about integration of vector valued functions.

By the chain rule we have that
$$\partial^2(u\circ A)=(\partial^2 u\circ A)[\partial A]^2+(\partial u\circ A)\partial^2 A$$
and because $\partial^2 A=0$ and $\partial Ax=A$ we found that
$$\partial^2(u\circ A)(x)=(\partial^2 u\circ A)(x)A^2$$
what also can be written as
$$\partial^2_k(u\circ A)(x)=\big(H_{u(A)}Ae_k|Ae_k\big)=\big(A^{-1}H_{u(A)}Ae_k|e_k\big)\tag1$$
where $H_{u(A)}$ is the hessian matrix of $\partial^2 u\circ A$. Hence
$$\begin{align}\big(A^{-1}H_{u(A)}Ae_k|e_k\big)&=\left(A^{-1}H_{u(A)}A^\bullet_k|e_k\right)\\&=\left(A^{-1}\sum_{j=1}^n [H_{u(A)}]^j_\bullet A^\bullet_k|e_k\right)\\&=[A^{-1}]^k_\bullet H_{u(A)} A^\bullet_k\\&=\sum_{j,\ell=1}^n[A^{-1}]^k_j[H_{u(A)}]^j_\ell A^\ell_k\tag2\end{align}$$
But the orthogonality tell us that $[A^*]^k_j=[A^{-1}]^k_j=A^j_k$ and
$$[AA^*]^k_\ell=A_\bullet^\ell[A^*]_k^\bullet=\sum_{j=1}^n[A^*]_k^jA_j^\ell=\sum_{j=1}^nA_j^kA_j^\ell=\delta_{k,\ell}\tag3$$
Then from the above we find that
$$\Delta (u\circ A)=\sum_{k=1}^n\big(H_{u(A)}Ae_k|Ae_k\big)=\sum_{k,j,\ell=1}^n(\partial_{j,\ell}u\circ A) A^j_kA^\ell_k\\=\sum_{j,\ell=1}^n(\partial_{j,\ell} u\circ A)\sum_{k=1}A^j_kA^\ell_k=\sum_{j,\ell=1}^n(\partial_{j,\ell} u\circ A)\delta_{j,\ell}=\Delta u\circ A$$
