This presentation describes the construction of an outer automorphism of group $S_6$. To begin with, we construct a homomorphism from $S_5$ to $S_6$, using transitive action. It is further argued that $$|\ker f| \le\dfrac{|S_5|}{6} = 20$$ And the question is, where does this inequality come from? Thanks for the help!
UPD: you can find detailed explanation here
If a group $G$ acts transitively on a set $X$ via $f\colon G\mapsto \operatorname{Sym}(X)$, then the image of $f$ has at least order $|X|$, hence the kernel is of index at least $|X|$.
A homomorphism $f\colon S_5\to S_6$ is equivalent to an action of $S_5$ on the set $X:=\{1,2,3,4,5,6\}$. If this action is transitive, then there is one orbit only, necessarily of size $|X|=6$, and thence (Orbit-Stabilizer Theorem):
$$6|\operatorname{Stab}(i)|=5!, \space\forall i\in X \tag 1$$
whence:
$$|\operatorname{Stab}(i)|=20, \space\forall i\in X \tag 2$$
Now, in general, for a $G$-action on a set $X$, the kernel of the equivalent homomorphism $f\colon G\to \operatorname{Sym}(X)$ is given by:
\begin{alignat}{1} \operatorname{ker} f &= \{g\in G\mid f_g=\iota_X\} \\ &= \{g\in G\mid f_g(x)=\iota_X(x), \forall x \in X\} \\ &= \{g\in G\mid g\cdot x=x, \forall x \in X\} \\ &= \{g\in G\mid g\in \operatorname{Stab}(x), \forall x \in X\} \\ &= \{g\in \operatorname{Stab}(x), \forall x \in X\} \\ &= \bigcap_{x\in X}\operatorname{Stab}(x) \\ \tag 3 \end{alignat}
Therefore, in our case, by $(3)$ and $(2)$:
$$|\ker f|=|\bigcap_{i\in X}\operatorname{Stab}(i)|\le 20 \tag 4$$
