I encounter the following problem:
Let $S^3$ denote all the $3\times 3$-real symmetric matrices. If $F:S^3\to S^3$ is a linear mapping, with $F(OAO^T)=F(A)$, for any $O\in\left\{\pmatrix{1&0&0\\0&1&0\\0&0&1},\pmatrix{0&0&1\\1&0&0\\0&1&0},\pmatrix{0&1&0\\0&0&1\\1&0&0}\right\}$, then can we conclude that $F(A)=\lambda A+\mu\mathrm{tr}(A)E_3$, for some constant $\lambda,\mu$? where $\mathrm{tr}$ is the trace and $E_3$ is the identity matrix.
After a try, I find that the condition is not enough, what natural condition could be add to make the conclusion holds?
The answer is negative.
Call the three permutation matrices your question as $I$ (sorry, I just can't stand denoting the identity matrix as $E_3$), $P$ and $Q$. For $i\neq j$, let $S_{ij}$ denote the matrix with a single one at the $(i,i)$-th entry and zero elsewhere, and let $S_{ij}$ be the matrix with ones at the $(i,j)$-th and $(j,i)$-th entry and zero elsewhere. Then \begin{cases} PS_{11}P^T = S_{22},\\ PS_{22}P^T = S_{33},\\ PS_{12}P^T = S_{23},\\ PS_{23}P^T = S_{31}. \end{cases} Hence the condition that $F(OAO^T)=F(A)$ implies that \begin{cases} F(S_{11})=F(PS_{11}P^T)=F(S_{22})=F(PS_{22}P^T)=F(S_{33}),\\ F(S_{12})=F(PS_{12}P^T)=F(S_{23})=F(PS_{23}P^T)=F(S_{31}). \end{cases} That is, $F(S_{11})=F(S_{22})=F(S_{33})=B$ and $F(S_{12})=F(S_{23})=F(S_{31})=2C$ for some real symmetric matrices $B$ and $C$. Hence the general form of $F$ is given by $$F(A)=(a_{11}+a_{22}+a_{33})B + 2(a_{12}+a_{23}+a_{31})C.$$
Consequently, it is not necessarily true that $F(A)=\lambda A+\mu\mathrm{tr}(A)I$ (in fact, if $F$ is of this form, $\lambda$ must be zero). For a counterexample, consider $B=C=S_{11}$ and $F(A)=(\mathbf{1}^TA\mathbf{1})S_{11}$, where $\mathbf{1}=(1,1,1)^T$.