Yes – there is a clean answer that simultaneously addresses your three questions.

Set-up and a standard description of Aut(Gn) For any group G, write Zi = Z(G) for its center. A standard and very useful fact is that Aut(Gn) is generated by:

• permutations of the n coordinates,
• componentwise automorphisms (α1,…,αn) with each αi ∈ Aut(G), and
• “elementary transvections” tij(φ) (i ≠ j) defined by tij(φ)(g1,…,gn) = (g1,…,gi φ(gj),…,gn), where φ: G → Z(G) is a homomorphism. This is an automorphism because φ(gj) ∈ Z(G).

Thus, to check that a subgroup K ≤ Gn is characteristic, it suffices to check invariance under these generators.

A simple criterion Let H ≤ G. Define T(H) := ⟨ φ(h) | h ∈ H, φ ∈ Hom(G, Z(G)) ⟩ ≤ Z(G).

Then the following are equivalent:

(1) H × H is characteristic in G × G. (2) H is characteristic in G and T(H) ≤ H. (3) Hn is characteristic in Gn for every n ≥ 1.

Sketch of proof:

• (3) ⇒ (1) is obvious.
• (1) ⇒ (2): From (1), invariance under the diagonal copy of Aut(G) in Aut(G × G) gives H characteristic in G. Invariance under each transvection t12(φ) forces φ(H) ≤ H for all φ: G → Z(G), hence T(H) ≤ H.
• (2) ⇒ (3): Invariance under coordinatewise automorphisms follows from H characteristic; invariance under permutations is clear; and invariance under every transvection tij(φ) follows from φ(H) ≤ H. Since these generate Aut(Gn), Hn is characteristic in Gn for all n.

Consequences and answers to your three questions

1. Strongest form: If H × H is characteristic in G × G, then Hn is characteristic in Gn for all n. Yes. In fact, as Marty Isaacs noted, the part “H characteristic in G” is redundant: H × H characteristic already implies H characteristic, and as just shown, it also gives the extra condition T(H) ≤ H that is precisely what is needed for all n.

2. Intermediate form: Is there a finite n0 to check? Yes: n0 = 2 suffices. In fact, H × H characteristic in G × G ⇔ Hn characteristic in Gn for all n ≥ 2. So checking n = 2 already decides the property for every n.

3. Weakest (finite test/characterization): There is a finite characterization in group-theoretic terms: Hn is characteristic in Gn for all n if and only if

• H is characteristic in G, and
• T(H) ≤ H, i.e., φ(H) ≤ H for every φ ∈ Hom(G, Z(G)).

For finite groups, this is an effective finite test (there are finitely many homomorphisms G → Z(G)). For finitely generated groups with finitely generated center, it also reduces to finitely many checks because Hom(Gab, Z(G)) is finitely generated; it suffices to test a finite generating set of homomorphisms. In full generality (arbitrary groups), this yields a precise algebraic criterion, though not always an algorithmic one.

Remarks and special cases

• If Z(G) = 1, then T(H) = 1, so every characteristic H automatically satisfies Hn characteristic in Gn for all n. In particular, the obstruction occurs only when Z(G) ≠ 1.
• In the abelian case, Z(G) = G, so the condition “φ(H) ≤ H for all φ: G → Z(G)” is exactly “H is fully invariant.” Thus, for abelian groups, the property “Hn characteristic in Gn for all n” is equivalent to “H is fully invariant,” agreeing with your aside (for finite abelian groups of odd order, every characteristic subgroup is fully invariant).
• Your original counterexample with G = Z8 × Z2 and its “weird diagonal” H shows that, without the assumption that H × H is characteristic, Hn need not be characteristic in Gn even for n = 2. The extra condition H × H characteristic forces the “transvection” invariance T(H) ≤ H that fails there.

This also aligns with your intuition: “positive” descriptions (verbal/fully invariant subgroups) are closed under all endomorphisms and hence under the transvections that appear in Aut(Gn); “negative” characterization often fails exactly at this transvection step.