For a finite group $G$, let $β(G)$ denote the minimum cardinality $|P|$ among finite posets $P$ whose automorphism group $\Aut(P)$ is isomorphic to $G$. While every finite group is realizable as the automorphism group of some finite poset, exact values of $β(G)$ are known only in special cases, most notably for cyclic groups. In this paper we prove that $β(\mathbb{Z}_2 \times \mathbb{Z}_4) = 14$; in particular, the product bound $β(G \times H) \le β(G) + β(H)$ is sharp in this case. The upper bound is realized by an explicit $14$-element poset $P_{14}$, whose automorphism group is computed by a height-function argument together with a rigidity analysis of its covering relations. The lower bound, which constitutes the substantive part of the proof, is established by a case analysis of the orbit decompositions of a hypothetical poset on at most $13$ points under a faithful $G$-action, organized according to the largest orbit size; in each case we construct an order-automorphism outside the given copy of $G$, contradicting $\Aut(P) \cong G$. Among non-cyclic groups, to our knowledge this is the first exact determination of $β(G)$ whose lower bound requires a structural analysis of this kind: for the other non-cyclic abelian groups of order at most $8$, namely $\mathbb{Z}_2 \times \mathbb{Z}_2$ and $\mathbb{Z}_2^3$, the value of $β$ is elementary. The arguments are closely adapted to the subgroup lattice of $\mathbb{Z}_2 \times \mathbb{Z}_4$.