Let $\Sigma_n$ be the symmetric group on $n$-letters. My question is can $\Sigma_n$ act non-trivially on a set of $k$ elements say $\{1,2,\dots ,k\}$ where $k<n$?
I can't think of any examples where it acts non-trivially. So, I am inclined to think it can't. But I am not sure about it. Any help would be appreciated.
An action of a group $G$ on a set with $k$ elements induces a homomorphism $\phi\colon G\to \Sigma_k$. If $\Sigma_n$ acts on a set with $k\lt n$ elements, the action induces a morphism $\Sigma_n\to\Sigma_k$. Because the map cannot be one-to-one from size considerations, the kernel is nontrivial. If you want the action to not be trivial, at least for $n\geq 5$ you need the kernel to be $A_n$. In that case, the action factors through $\Sigma_n/A_n\cong C_2$. We can do this with no fixed points if $k$ is even; if $k$ is odd, though, the action must have a fixed point.
For example, we can let $\Sigma_5$ act on $\{1,2,3,4\}$ by letting odd permutations act as $(12)(34)$, and even permutations act trivially. But $S_5$ cannot act on $\{1,2,3\}$ without having at least one fixed point.
For $n=4$ the action can also factor through $\Sigma_4/V$, where $V=\{e,(12)(34),(13)(24),(14)(23)\}$; this is a group of order $6$, isomorphic to $S_3$, and so $\Sigma_4$ can act on $\{1,2,3\}$ with no point being fixed by the action. For $n=3$, the only nontrivial normal subgroup is $A_3$, so it works as it does for $n\geq 5$.
In summary:
It cannot be done for $n=2$.
For $n\gt 2$, $n\neq 4$, there are actions of $\Sigma_n$ on $\{1,2,\ldots,k\}$ with $k\lt n$, $k$ even, with no points fixed by the action: just let odd permutations act as a product of transpositions that permute all points.
For $n\gt 2$, $n\neq 4$, any action of $\Sigma_n$ on $\{1,2,\ldots,k\}$ with $k\lt n$, $k$ odd, has a fixed point.
For $n=4$, $\Sigma_4$ can act nontrivially on a set with two elements (let odd permutations exchange the two points, even permutations fix both points), and on a set with $3$ elements (acting as $S_3$ via $\Sigma_4/V$.