Suppose there is not finitely-generated group $G$ and a finite set $X$. Then it is impossible that the set of all possible maps $f : X \to G$ is finitely generated.
Proof. If $H = \{ f : X \to G \}$ were finitely generated say by $H = \langle f_1, \dots, f_n\rangle$ with $n \in \Bbb{N}$. Then for $h \in H$ let $w(h) = $ the smallest word size such that $h = $ a product of $w(h)$ $f_i$ inverses or direct usages. $w(1) = 0$. Clearly $w(gh) \leq w(g) + w(h)$.
Then ?
Hint: If $X\neq\emptyset$, then there is a surjective homomorphism $G^X\rightarrow G$. Use this to show that if $G^X$ is finitely generated, then so is $G$. (Note: $X$ is not required to be finite)