The question is asking me to prove that $f(a \circ b) = f(a) \circ f(b)$.
This I believe is referencing our previous proof which tells us: Assume $g:x \rightarrow y $ is a bijection and for an $a \in S(X)$ set $f(a)= g \circ a \circ g^{-1}$ and then I proved $f$ is a bijective function from $S(X)$ to $S(Y)$.
We haven't been given any background on homomorphisms so I'm a little unsure on how to proceed.
Note that $f(ab)=gabg^{-1}=ga1bg^{-1}=gag^{-1}gbg^{-1}=\cdots$?