This is a cross-post from math.se, because I did not get any answer there:
Write $[n]:=\{1,\ldots,n\}$. For a partition $\lambda\vdash n$, I will write $[\lambda]$ for the Specht module that corresponds to $\lambda$, i.e. the complex vector space spanned by all standard tableaux of shape $\lambda$.
Consider the complex vector space $V_k$ which has as a basis the $k$-element subsets of $[n]$. The symmetric group $S_n$ acts on $V_k$ in the was that $g\in S_n$ maps a basis vector $A\subseteq[n]$ to the basis vector $g(A)=\{ g(i) \mid i\in A \}$. Let us assume $k\le \frac n2$. It is known that $$V_k = \bigoplus_{j=0}^k~ [n-j,j]$$ is the decomposition of $V_k$ into irreducibles. See, for instance, Example 7.18.8 in Stanley's book Enumerative Combinatorics (Volume 2).
Question: Given any $j\le\frac n2$ and a standard tableau $T$ of shape $\lambda=(n-j,j)$, I would like to know how to express the corresponding basis vector $v_T$ of $[\lambda]$ in the canonical basis of $k$-element subsets of $[n]$.
Thoughts: For $k=1$, this is rather easy. For the only standard tableau $T$ on $\lambda=(n)$, the vector $v_T=\sum_{i=1}^n \{ i \}$ is just the sum of all $1$-element subsets of $[n]$. On the other hand, for a standard tableau $T$ of $(n-1,1)$, let $a_T$ be the number in the single box in row $2$. Then it can be checked that $v_T=\{1\}-\{a_T\}$. However, I already have trouble with $V_2$. There is an obvious $S_n$-invariant inclusion \begin{align*} \phi_1: V_1 &\longrightarrow V_2 \\\\ \{i\} &\longmapsto \sum_{j\ne i} \{ i,j \} \end{align*} and the $[n-2,2]\subseteq V_2$ must be isomorphic to $V_2/\mathrm{im}(\phi_1)$, but I didn't get much further.
