I don't think I understand the question right, because it seems trivial.
Lets mark one set of flags A,B,C the second set a,b,c with A and a one color, B and b the next color, C and c the remaining color (colors respectively red, green, blue, or whatever permutation).
Arrange the first set A,B,C into an equilateral triangle. Mark the middle of each side, and put the remaining flag of the opposing corner color there: So the main big triangle A-B-C now has sides with middle points A-c-B, B-a-C, C-b-A.
That solves the situation because
All triangles that can occur are equilateral triangles; all non-equilateral triangles -- say aBb --- have two the same colors.
This is provable by contradiction because
ONE or more original/first-set vertices are involved (otherwise it's the inner equilateral triangle abc you deal with), assume A (re-labeling makes it hold for the others, like I relabeled colors anyway to start); and ONE or more second-set vertices must be involved (otherwise it's the original equilateral triangle ABC you deal with), assume this is b (again, b by relabeling if necessary). Now the third vertex must be c or C; it cannot be C as that's collinear/ not a triangle, so we're left with Abc which is equilateral.
Or did you want to allow for "collinear triangles" (which are not triangles)? EDIT: Bof, if you allow for those (which by now, reading http://mathworld.wolfram.com/Triangle.html it satisfies that definition for a limit case where the angles tend to 0,0,180), then any collinear points (A,B,C) with B between A and C form a (degenerate) triangle with the angles at A and C being zero, at C it's 180 (or pi). But according to the definition I'm following here --- http://mathworld.wolfram.com/SimilarTriangles.html --- any three collinear points gives a similar triangle because it's the angles of the vertices NOT the length ratios of the sides. So ABC collinear whether a limit of ABC isosceles (so |AB|=|BC| in the limit) or not, that's all similar because the same angles. So arrange all six in any order on a single line, and Pheno Menon's question is answered by example. [Note that then they're also inversely similar as you can reach the limit from the mirrored triangle.]
If now somebody invokes some further overlooked rule why they would not be similar, and only collinear triplets at the same ratio |AB|:|BC| are similar, then clearly the arrangement with all six collinear is ruled out. Which means
there cannot be any collinear triplets, and no equilateral triangles either; then you start with a (non-degenerate) triangle ABC and see that either abc are all in the interior of ABC, all in the exterior (which is the same case as previous but reversed), or one in the interior two exterior, or the reverse. Here it's quick to see they cannot give you the same triplets of angles.