The number of edges in Kn is $\frac{n(n-1)}{2}$ so it's clear that either (a) $n$ is a multiple of three or (b) $n-1$ is a multiple of three.
The proof doesn't have to be extremely rigorous, I just want to get the idea OR know how to build the three isomorphic graphs.
For the case $n=3m$, you can partition the vertex set $V$ into $3$ blocks: $V_1, V_2, V_3$, such that $|V_i|=m$ for all $i$. For each $V_i$, you can then look at $E|V_i$, that is, the set of edges whose endpoints are both in $V_i$- this is just a copy of $K_m$. However, there are still some edges left, namely the edges with endpoints in different blocks: for example, the set of edges with one endpoints in $V_1$, and the other in $V_2$. We can call this set of edges $E_{v_1, V_2}$, and all of the remaining edges are partitioned among three such blocks, which we call $E_{V_1, V_2}, E_{V_1, V_3}, E_{V_2, V_3}$. Each such block is isomorphic to the bipartite graph $K_{m,m}$. Since we originally looked at three disjoint blocks of edges, each isomorphic to $K_m$, ($E|V_1, E|V_2, E|V_3$), and the remaining edges were partitioned among three blocks, each isomorphic to $K_{m,m}$, ($E_{V_1, V_2}, E_{V_1, V_3}, E_{V_2, V_3}$), we can then pair each block of the first kind with a block of the second kind, say $E|V_1$ goes with $E_{V_2, V_3}$, take their union, and get three disjoint blocks of edges that are pairwise isomorphic.
For the case $n=3m+1$, you can do the same thing, but with an extra vertex. Can you see how to extend the previous construction with this extra vertex?