If there is an orthogonal similarity between symmetric matrices $A$ and $B$ by having $B=OAO'$ for an orthogonal matrix $O$ ($'$ is transpose) we infer $A$ and $B$ are having identical eigenvalues (since $A$ is symmetric there is orthogonal $O_1$ satisfying $O_1\Lambda O_1'=A$ where $\Lambda_1$ is diagonal and $B=OO_1\Lambda O_1'O'\iff O_1'O'BOO_1=\Lambda$ diagonalizes $B$).
If $A$ and $B$ are symmetric and are having identical eigenvalues is there always an orthogonal $O$ satisfying $B=OAO'$?
Yes there is, if $A$ and $B$ are symmetric and have the same eigenvalues then there exists two ortogonal matrices $O_{A}$ and $O_{B}$ and a diagonal matrix $\Lambda$ such that the following equations holds:
$$A=O_{A}\Lambda O_{A}^{T}$$ and $$B=O_{B}\Lambda O_{B}^{T}$$ this implies $$\Lambda=O_{A}^{T}A O_{A}$$ and $$\Lambda=O_{B}^{T}B O_{B}$$ so we have $$O_{A}^{T}A O_{A}=O_{B}^{T}B O_{B}$$ $$A=O_{A}O_{B}^{T}B O_{B}O_{A}^{T}$$, but $O_{A}O_{B}^{T}$ is an ortogonal matrix becuase is product of two ortogonal matrices.