It is well known that the non abelian exterior square of $S_n$ is again a group of order $n!~(n>2)$ e.g for $n=5, ~ S_n\wedge S_n\cong \text{SL}(2,5)$ (https://groupprops.subwiki.org/wiki/Exterior_square_of_a_group). My question is that:
Is this possible that for some $n$, $S_n\wedge S_n\cong S_n$? I have seen that it is not possible for $n<7$, but above $7$, I do not have any information.
Please help me to solve this problem.
Looking at the link you provided, there is a surjective morphism $S_n\wedge S_n\to [S_n,S_n]=A_n$, so $A_n$ is a quotient of $S_n\wedge S_n$. But $A_n$ is not a quotient of $S_n$ since, for $n>2$, $S_n$ does not have a normal subgroup of order $2$.