While reading this paper, I came across following argument (in Lemma 5):
Consider the least $k$ such that $f (I_k) = I_j$ and $j < k$. Clearly $k > 1$ and $f(I_{k−1})
= I_m$ for some $m > k − 1$. We consider three cases: (a) $j = k − 1$ and $k = m$;
(b) $j ≤ k − 1$ and $m > k$; (c) $j < k$ and $m ≥ k$. We claim that the second two cases
are impossible. The arguments are similar in both cases so we consider case (b).
In this case, we have that $f (s_{k−1}) \in [r_m, s_m] = I_m$, so that $f(s_{k−1}) ≥ s_k$ and, similarly,
$f (r_k) ≤ s_{k−1}$. This implies that the image of the interval $L = [s_{k−1},r_k]$ under $f$ contains the interval $I_k$. But no point of $I_k$ is mapped into $L$, so some subinterval of $L$ does not contain any periodic points, which is a contradiction. So the only possibility is case (a).
I was able to follow the argument till here. However, I'm not able to understand the next line.
Since $f$ permutes the intervals $I_i$, for $i ≤ n$, (a) implies that $n = 2$.
How do we conclude that $n=2$?