Sonny-
Yes it is legal but I think you may have the cart before the horse. When you talk about these different methods of utilizing your cable tray, you have a serious impact on the ampacity rating on your cable. There are a few other factors to consider in addition to the method you use.
From my experience, cable "bundling" is the most effiecient method to use. If you look at how it's laid out, each bundle is spaced 2 on bottom, 1 on top leaving a space for 2.15 outside diameter in between. In my applications, we also include a fully rated neutral. So if I took the 2 cables off the top, I could fit them essentially in between them. The difference is in the cable ampacity rating that you get. I will have to make some assumptions here that may be different for your situation.
Ambient temp = 40C
Transformer connections and/or MCC rated at least 75C
All the cables in the vented or ladder cable tray (uncovered) are for the transformer output and the same length.
392.11 lists 3 methods for cabling and how to determine the ampacity for each:
1. Single layer of cables, no space in between.
1000 MCM found in Table 310.17 (Free Air rating) = 935 A
935 A x .88 (temp derate) = 822.8 A per cable
822.8 A x 75% = 617.1 A per cable
617.1 A x 5 = 3085.5 A total.
2. Single layer of cables maintaining 1 cable space between.
1000 MCM found in Table 310.17 (Free Air rating) = 935 A
935 A x .88 (temp derate) = 822.8 A per cable
(Good news: no further derating required. Bad news: Your cable tray is 2X the size!)
822.8 A x 5 = 4114 A (I think this would be over-kill for your situation. And given the price of copper today. . .

)
3. Bundling one cable of each phase together.
1000 MCM found in Table 310.20 (3 insulated cond'rs. on messenger) = 748 A
748 x 5 = 3740 A
This method best utilizes the space on the cable tray AND has the best ampacity. So you might be able to drop some of those cable and save $$$$
One last thing: If you do use that last method, you can calculate the minimum size of the cable tray = [Outside Diam. of conductor] x 2.15 x 2 x [number of runs]. AND THEN, bounce that off the Table 392.10(A). Some of the other in this forum might have a more effiecient and technically correct method of calculating the size but this will ensure that you have enough spacing in between the cables. I'd rather have 2 extra inchs that being 1 inch short.
Let me know what you decide.
Edit to note that two replies were sent since I started--I take too long!