Intuition strongly suggests that there exist  $\left\lfloor\frac{\binom{n}{k}}{\lfloor\frac{n}{k}\rfloor}\right\rfloor$ independent sets  in the complement of a Kneser graph  each having $\lfloor\frac{n}{k}\rfloor$ vertices in it. Is this true. If true how to establish it?

A construction of such a set of cliques in the Kneser graph $K(6,2)$ is as follows:
$$(12)(34)(56)$$
$$(13)(25)(46)$$
$$(14)(26)(35)$$
$$(15)(24)(36)$$
$$(16)(23)(45)$$
Thus, in this example we have $5$ triangles in the Kneser graph $K(6,2)$ which correspond to an equitable $5$ coloring of the complement graph $\overline{K}(6,2)$. Can such a construction be always done? I think this is related to the number of order $2$ elements in the symmetric group of order $n$. Thanks beforehand.