I've found all the $3\times 3$ matrix representations of the permutation representation of $S_3$ with respect to the standard basis $\{e_1,e_2,e_3\}$, but I'm then tasked with finding a different basis such that all the matrices are of the form: $$\begin{pmatrix}1&0&0\\0&*&*\\0&*&*\end{pmatrix}$$
Given my current understanding of the problem, this seems impossible, as if $\{b_1,b_2,b_3\}$ is our different basis, then we need $\sigma b_1=e_1$ for all $\sigma \in S_3$, which is just not possible. Clearly I don't understand something, and would appreciate help in realising exactly what it is that I have misunderstood.
Hint: If $(b_1,b_2,b_3)$ is a basis with respect to which all the permutations $\sigma$ are represented by matrices of the written form, then $\sigma(b_1) = b_1$ for all $\sigma \in S_3$ (not $\sigma(b_1) = e_1$!). In other words, $b_1$ should be a common eigenvector for all the permutations so that when $\sigma$ acts on $b_1 = (x,y,z)^T$ by permuting the coordinates, the vector $b_1$ stays invariant. Can you think of such a vector?
In addition, the subspace spanned by $(b_2,b_3)$ should be invariant under all the permutations. How can you complement $\operatorname{span} \{ b_1 \}$ in order to achieve it?