For $n\in \mathbb N$ consider the symmetric group $S_n$ as the group of all bijections from $\{1,\ldots,n\}$ onto itself. Then the following is impossible: There exists a subgroup $H$ of $S_n$ with $|S_n:H|=n$ and $H$ does not fix any element of $\{1,\ldots,n\}$.
How should I approach ?
My effort: Consider the action of $H$ on $\{1,\ldots,n\}$. Note that there is no fixed points. Every orbit has $(n-1)!$ elements. If there are $k$ orbits then $k\cdot(n-1)!=n$, i.e., $k=\frac{n}{(n-1)!}$. For $n>2$, $k$ fails to be an integer.
Sorry my arguments are wrong, orbits are not necessarily of same size.
As basically pointed out by Derek Holt in the comments, if $H < S_n$ is a subgroup such that $[S_n : H] = n$, then the left coset action on $H$ gives you a homomorphism $\varphi: S_n \rightarrow S_n$.
Show that $\varphi$ is an automorphism.
If $H$ is transitive, show that $\varphi$ must be an outer automorphism.
Conclude that $n = 6$.
Then at this point I suppose you have to look at $S_6$ more closely to conclude that the only possibility is $H = PGL(2,5)$. Some hints are in Dixon and Mortimer, Chapter 2, Exercise 2.9.8.