Consider the set $S = \{A, B\}$ and the group $G = \{e_G, \varphi\}$ with operation composition (where $e$ is the identity map and $\varphi$ maps $A$ to $B$ and $B$ to $A$). This is the permutation group on $S$ (that is, a group whose elements are the bijective mappings from $S$ to itself).
Given $\alpha$, a map from $S$ to $G$, why is it impossible for $G$ to be a free group? That is: If $\alpha'$ is a map from $S$ to some other group, $G'$, why is there not a unique homomorphism, $\mu$, from $G$ to $G'$ such that $\alpha' = \mu \circ \alpha$?
I'm asked to prove that the only group which can be both a free group and a permutation group is the group with a single element (the identity), but I am not convinced this is true. As far as I can tell, if we map the elements of $S$ to some $G'$, there's only one possibility when mapping elements of $G$ to $G'$, and said possibility would have to preserve composition within $G'$.
The first thing I think is: "Well, something in $S$ is necessarily mapped to $e_G$, and the only way $\mu$ can be a homomorphism is if $\mu(e_G) = e_{G'}$. Thus, the only way $G$ can be a free group is if something in $S$ mapped to $e_{G'}$ in the first place—which is not a constraint we can assume."
... but the group w/ only a single element has the identity as its element, so that assumption would be made anyway.