Consider the vector space: $$ \mathbb{V} = \{ \ A \in M_{3 \times 3}(\mathbb{R}) \ | \ A^{T} = A \mathrm{\ and\ }\mathrm{Tr}(A) = 0 \} $$
As well as the Lie Group: $$ O(3) = \{ \ A \in M_{3 \times 3}(\mathbb{R}) \ | \ A^{T}A = A A^{T} = \mathbb{I} \} $$
Define the group action $\Phi : O(3) \times \mathbb{V} \to \mathbb{V}$ as $\Phi(A,X)= \Phi_{A}(X) = AXA^{T}$. I am to determine the group orbit$s$ of this action.
The way I am understanding this, is that I am to examine the structure of the set $$ S := \{ \ \Phi_{A}(X) \in M_{3\times 3}(\mathbb{R}) \ | \ A \in O(3), X \in \mathbb{V} \ \} $$
I know that symmetric matrices are orthogonally diagonalizable, which means that I can write $X = P D P^{T}$ for any $X \in \mathbb{V}$ (where $P \in O(3)$ and $D$ is diagonal and traceless).
This tells me that $\Phi_{A}(X) = AXA^{T} = APDP^{T}A^{T} = \Phi_{AP}(D)$, where $AP$ is just another matrix in $O(3)$. This tells me: $$ S = \{ \ \Phi_{A}(D) \in M_{3\times 3}(\mathbb{R}) \ | \ A \in O(3), D = \mathrm{diag}(a,b,c) \mathrm{\ with\ }a+b+c=0 \ \} $$
So the orbits of the whole vector space are exactly the orbits of the diagonal matrices in this subspace.
Can I narrow this down further?