This was said in Prof. Balakrishnan lecture 19 on quantum mechanics for the case of exchange symmetry, but he showed no reason why.
For example, the system corresponding to two spin $\frac{1}{2}$ systems has a singlet state, which is antisymmetric, and three triplet states, which are symmetric. Apparently this has to be always the case. I am also assuming that a multiplet means a set of states with the same total angular momentum (infering from the spin $\frac{1}{2}$ case), although Wikipedia seems to phrase it in more group-theoretic terms with which I'm not yet familiar.
Best Answer
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. 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. Balakrishnan'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.