Here is exercise 1.1.17 from Introduction to Graph Theory by Douglas B. West
1.1.17. Prove that $K_n$ has three pairwise-isomorphic subgraphs such that each edge of $K_n$ appears in exactly one of them if and only if n is congruent to $0$ or $1$ modulo $3$.
I am trying to formalize a proof for this exercise, but struggle simplifying my arguments. I would appreciate if you could pinpoint where I should be clearer or where I am wrong. Here is what I've sketched so far.
Sketch
The number of edges for any given n-vertex clique $K_n$ is $n(n-1)/2$.
If $3 | n+1$ then $3\nmid n(n-1)$. Therefore, $e(K_n)$ is divisible by three if and only if $n+1$ is not divisible by three.
To us this means that the edge set of a n-vertex clique can be partitioned into three equally sized subsets of edges given that $n(n-1)$ is divisible by three.
First, consider the case where $n$ is divisible by three. Now, imagine we are reconstructing a clique with three pairwise isomorphic subgraphs. And observe how the following algorithm induces $K_n$.
Draw three copies of $K_{n/3}$ which are disjoint from one another.

Pair every copy and add an edge within a pair if two vertices are not adjacent to each other.

And so we've induced $K_n$. Because copies are isomorphic with respect to one an other, then all pair of copies must also be isomorphic with respect to one another. Therefore, in the case where $n$ is divisible by three then the claim in 1.1.17 is shown to be true.
The reasoning for the case where $n+1$ is divisible by three is corollary to this framework so I will not highlight it.
There is a simpler way to do this. You can partition, iff $n \equiv_3 0$ or $n \equiv_3 1$, the set $E(K_n)$ into $3$ subgraphs where each of the $3$ subgraphs consists of
For $n \equiv_3 0$:
For $n \equiv_3 1$: