the main thing to do is to prove how the stabilizer changes for elements in teh same orbit (the formula that I wrote above). And of course the stabilizer for $\overline{1}$ in $G/H$ is $H$
but then i dont get the question im solving: we are asked to show that $k=2^n$ is the smallest $k$ s.t. there exists an embedding of $Q_{2^n}$ into $S_k$
the hint is to identify the elements of order 2 in $Q_{2^n}$
in $S_m$ there are a lot of elements of order $2$ for every given $m$ (in general)
@user123 so let $z$ be the unique element of order $2$ in $Q_{2^n}$. By Cauchy's theorem, every non-trivial subgroup of $Q_{2^n}$ contains $z$. Suppose that we have an embedding of $Q_{2^n}$ into $S_k$ for $k < 2^n$, i.e. a faithful action on a set of $k$ elements for $k < 2^n$
let $\alpha:Q_{2^n} \to S_k$ be that embedding
let $X$ be the $k$-element set on which $S_k$ (and hence $Q_{2^n}$) acts. You know that $\mathrm{ker}(\alpha)= \cap_{x \in X} \mathrm{stab}_G(x)$
do you see how this helps together with the fact that $z$ is contained in every nontrivial subgroup of $Q_{2^n}$?
Now here's the part where $k < 2^n$ is crucial: $|\mathrm{Orb}_G(x)| \leq k < |G|$, since $G$ acts on a $k$ element set, so $|\mathrm{stab}_G(x)|=1$ is impossible
thus $\mathrm{stab}_G(x)$ is a non-trivial subgroup of $G$ and thus it contains $z$
and so $z$ must also be contained in $\cap_{x \in X} \mathrm{stab}_G(x) = \mathrm{ker}(\alpha)$
