Past exam of the mathematics course of the University of Cambridge / 2024 / ii / Paper 4 / 4J / c

False. Given any $D$, adjoin arbitrarily many unreachable states whose transitions remain among those new states, choosing their accepting status consistently away from the embedded copy. This produces infinitely many pairwise nonisomorphic finite deterministic automata $D^{\prime }$ containing an injective homomorphic copy of $D$.

True. In fact any fixed finite $D$ works. If $D^{\prime }\le D$, then $D^{\prime }$ has at most $|Q|$ states. For a fixed finite alphabet, there are only finitely many transition tables, initial states, and accepting subsets on at most $|Q|$ labelled states, hence only finitely many isomorphism classes.