Haha, that makes this the 50th version of FF4kster I've downloaded on this computer alone. I am glad to see that this problem appears to be fixed now.

I'm glad it turn out to be quickly fixed (<15 mins!!) - happy editing!

I in no way guarantee this to be the answer - but I think it might at least shed a little light on the Giant problem. I pulled out the individual tiles for three sets: Giant, Whale, & Tower - I put each tile side by side (rows of 3) so one row per hex assignment so I could do the only thing I know how to do: patterning!

I hit the Ship/Airship tilesets the same way - the result is more straightforward only comparing 2 rather than 3. Comparing the two tilesets they can be (mathematically) be split into subsets:

Airship = (A) + (C) & Ship = (S) + (C)

Where: (A) are tiles unique to the Airship set, (S) are tiles unique to the Airship set, & (C) are common to both

Specifically the common subset is comprised of the tiles of the indices:

{01, 02, 03, 04, 05, 06, 07, 08, 09, 0A, 0B, 0C, 0D, 0E, 0F, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1A, 1B, 1C, 1D, 1E, 1F, 20, 21, 22, 23, 24, 25, 26, 27, 28, 2A, 2B, 2C, 2D, 2E, 30, 32, 33, 35, 36, 37, 38, 39, 3A, 3B, 3C, 3D, 3E, 3F, 53, 54, 55, 56, 57, 58, 59, 5A, 5B, 5C, 5D, 5E, 5F, 6F, 7C, 7D, 7E}

& The individual sets are indexed:

{29, 2F, 31, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 4A, 4B, 4C, 4D, 4E, 4F, 50, 51, 52, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 6A, 6B, 6C, 6D, 6E, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 7A, 7B, 7F}

Of the 127 indices 76 are shared, and 51 unique. At the very least there must be some way the game differentiates referencing the unique subsets vs the common ones; but I don't yet know how.

Could you give me an idea of how for the tilesets already mapped out you build those complete tilesets from the metatile pool: that's something I don't know the specifics of (but I think I finally grasped the concept)

If I were to continue my analysis further the next step would be to split each tile in the sets in question into their 4 individual metatiles and then construct each minimum set, taking into account for each right-rotation:

...stupid combinatorics