Define $\mathbf{J}=\mathbf{S}_1 + \mathbf{S}_2$ where $\mathbf{J}$ by definition is the total angular momentum to two identical, say spin-half, angular momenta. Define $\sigma$ to be the operation that exchanges the two spins, i.e., it denotes the operation: $1\leftrightarrow 2$. Then, it is easy to see that $\sigma$ commutes with $\mathbf{J}$ as $\mathbf{S}_1 + \mathbf{S}_2=\mathbf{S}_2 + \mathbf{S}_1=\mathbf{J}$. (The argument can be extended to the case of the addition of $N$ identical spins where $\sigma$ gets replaced by any pairwise exchanges which generates the permutation group in $N$ elements.)
Coming back to the case of two spins, it follows that we can simultaneously diagonalise $J^2,J_z$ and $\sigma$. The action of $\sigma$ must necessarily take an eigenstate of $J^2,J_z$ to one with the same $J^2,J_z$ eigenvalues. But the addition of two angular momenta says, that the multiplicity of $J^2$ eigenvalues is one. This is not true when you add more than two spins. It follows that for each given value of $J^2$, the multiplet has a definite $\sigma$ eigenvalue. 976
Since $\sigma^2=1$, its eigenvalues must be $\pm1$.
However, neither $S_{1z}$ or $S_{2z}$ commute with $\sigma$. Thus, we have $$ \sigma\ |\ell,m_1,\ell,m_2\rangle = |\ell,m_2,\ell,m_1\rangle\ , $$ in the notation where $\ell(\ell+1)$ is the $S_i^2$ eigenvalue and $(m_1,m_2)$ are the eigenvalues of $(S_{1z},S_{2z})$ respectively. So in general, the action of sigma is not diagonal, unless $m_1=m_2$. Coming to Prof. <PERSON>'s example. He considered two spin half-particles. The $(j=1,m=\pm1)$ states only arise when $m_1=m_2=\pm\tfrac12$ and from the above argument must be symmetric.However, the $(j=1,m=0)$ and $(j=0,m=0)$ states arises as linear combinations of $m_1=-m_2$ state which are not diagonal. However, since the $(j=1,m=0)$ state can be obtained by using the lowering operator $J_-$ (which commutes with $\sigma$), it will have the same $\sigma$ eigenvalue as the state it was constructed from, i.e., $(j=m=1)$, which is symmetric. The singlet state must be (i) orthogonal to the triplet $m=0$ state and must necessarily be an eigenstate of $\sigma$ (I can give the precise argument if you wish) (ii) antisymmetric.
For the sum of N identical spins, all one can say is that there exists a basis of multiplets organised/labelled by the representations of the permutation group (given by Young diagrams) in addition to the $J^2$ eigenvalue. 66