Are all $n-1$ edge colorings of $K_n$ isomorphic?

Question

Baranyai's theorem tells us that a $K_n$ can be colored with $n-1$ colors whenever $n$ is even; the handshaking lemma makes it easy to show that is not possible when $n$ is odd. It's not hard to verify that for $n=4$ or $n=6$ this coloring is unique up to isomorphism (where we're allowed to reorder vertices and relabel colors). Is this always true? Are there larger even $n$ where are are multiple distinct colorings?

Answer 1

If $n$ is even, then by the result in this answer, every $(n-1)$-edge-coloring of $K_n$ corresponds to a symmetric $n-1$ by $n-1$ Latin square.

The number of symmetric Latin squares grows much faster than the number of symmetries of an $(n-1)$-edge-coloring, so for large $n$ there will be many possible colorings.

For $n=8$, it is not too hard to find an example of two non-isomorphic edge-colorings. On the one hand, there's the standard construction (shown below) which involves rotating the same perfect matching through multiples of $\frac{2\pi}{7}$ radians. You can check that in that construction, the union of any two perfect matchings is a Hamiltonian cycle. (Because of the rotational symmetry, there are only three cases to consider.)

On the other hand, if you start an edge-coloring by taking two perfect matchings whose union is two $4$-cycles, it shouldn't be hard to complete that to a $7$-edge-coloring of $K_8$. I did that in a fairly arbitrary way, and got the coloring below:

Because the property "Are there two colors whose union is not a Hamiltonian cycle?" is preserved by isomorphism, these two colorings are not isomorphic.