I am currently looking at the eigenvalue problems of the Laplace-Beltrami operator. Let $(M,g$) be a smooth oriented Riemann manifold. I am investigating the eigenvalue problem of the Laplace-Beltrami operator on $M$. The Laplace-Beltrami operator is given by $$ \Delta f = \sum_{i,j=1}^{n} \frac{1}{\sqrt{|g|}}\partial_{i}\left(\sqrt{|g|}g^{ij}\partial_{j}f \right),$$ where $g:=\operatorname{det}(g_{ij})$ and $g_{ij}$ are the components of the metric tensor and $g^{ij}$ are the entries of the inverse metric tensor.
I want to show (if its true) that the Helmholtz-equation i.e., the eigenvalue problem of the Laplace Beltrami operator, \begin{equation} \Delta u = \lambda u \qquad \qquad \text{in } M \end{equation} and, \begin{equation} u = 0 \qquad \qquad \text{on } \partial M \end{equation} $$ $$ is rotation invariant, i.e., if $(\lambda, u(x))$ is a solution to the Helmholtz equation in $M$ then $(\lambda, v(x))$ with $$ v(x) = u(Qx)$$ for some unitary matrix $Q \in \mathbb{R}^{n \times n}$is a solution to the Helmholtz-equation on $\widetilde{M}=\{ y|y=Qx, x \in M\}="QM"$.
Intuitively the statement makes sense since the Laplace-Beltrami operator only takes the local geometry into account. Thus, I would expect that the rotated eigenfunctions should be a solution to the rotated problem. I managed to prove the same statement for the classical Laplace operator. However, since in the Laplace-Beltrami operator both $g$ and $g^{ij}$ non-trivially depend on $x$, straight-forward calculations quickly became unfeasible. I also could not find a proof or counterexample in literature.