Show that Laplace's equation $\Delta u=0$ is rotation invariant

A function $u \in C^2(\Omega)$ is harmonic iff it statisfies the mean value propery. Then note that for any $B_r(x) \subset \Omega$ there holds $Q(B_r(x))=B_r(Qx)$. Then:
$$ \int_{B_r(x)}u(Qx)dx=\int_{B_r(x)}u(Qx)|\det(Q)|dx=\int_{Q(B_r(x))}u(x)dx=\int_{B_r(Qx)}u(x)dx=u(Qx) $$

Another way to see this is to resort to indices:

Let $O$ be the orthogonal matrix. Then

\begin{align} \Delta v & = \sum_{i=1}^n \frac{\partial ^2}{\partial x_i^2} u(Ax) \end{align}

Define $y_k = \sum_j A_{kj}x_j$. Note that $$ \frac{\partial v}{\partial x_i} = \sum_{j=1}^n \frac{\partial u}{\partial y_j}\frac{\partial y_j}{\partial x_i} = \sum_{j=1}^n \frac{\partial u}{\partial y_j} a_{ji}$$

Then we have \begin{align} \Delta v & = \sum_{i}\frac{\partial}{\partial x_i} \left(\sum_{j=1} \frac{\partial u}{\partial y_j}a_{ji}\right)\\ & = \sum_{ij}a_{ji}\frac{\partial }{\partial x_i}\frac{\partial u}{\partial y_j} \\ & = \sum_{ij}a_{ji}\sum_{k}\frac{\partial u}{\partial y_j \partial y_k}\frac{\partial y_k}{\partial x_i} \\ & = \sum_{ij}a_{ji}\sum_{k}\frac{\partial u}{\partial y_j \partial y_k}a_{ki}\\ & = \sum_{jk}u_{y_j,y_k}\sum_i a_{ji}a^T_{ik}\\ & = \sum_{j} u_{jj} = 0 \end{align}

Since $\sum_i a_{ji}a^T_{ik} = 1$ if and only if $j=k$ by orthogonality.