Show that isotropic function S(A) and A have same eigenvectors

Question

Given $\boldsymbol{A}$ is a positive definite, symmetric second order tensor and $\boldsymbol{Q}\boldsymbol{S}(\boldsymbol{A})\boldsymbol{Q}^T = \boldsymbol{S}(\boldsymbol{QAQ}^T)$ $\forall \boldsymbol{Q}$ which are orthogonal, then each eigenvector of $\boldsymbol{A}$ is also an eigenvector of $\boldsymbol{S}(\boldsymbol{A})$.

My specific question: If the orthogonal tensor $\boldsymbol{Q_1}$ diagonalizes $\boldsymbol{A}$, then $\boldsymbol{Q_1}\boldsymbol{A}\boldsymbol{Q_1}^T = \boldsymbol{D}$, so that

$\boldsymbol{Q_1}\boldsymbol{S}(\boldsymbol{A})\boldsymbol{Q_1}^T = \boldsymbol{S}(\boldsymbol{D})$.

Can I conclude from this that $\boldsymbol{Q_1}$ diagonalizes $\boldsymbol{S}(\boldsymbol{A})$ and hence the eigenvectors are the same ?

EDIT 1: I attach an image showing the standard proof. I understand it, but I would like to know if mine is wrong. EDIT 2: Changed my symbols to match that of the stated lemma.