if ker(φ) = {e}, φ must be injective, which would imply S3 and Z6 were isomorphic. why can't this be the case?

suppose ker(φ) = A3. then φ(S3) must be isomorphic to S3/A3, which has order 2. what are all the subgroups of Z6 of order 2?

can you see a possible way to define this homomorphism, based on the parity (even/odd) of an element in S3?

finally, you are overlooking another possiblity for ker(φ), which produces a rather dull homomorphism.....