While answering this question on common eigenvectors I wondered about the natural generalisation. Although I can deal with small cases I cannot find the general answer.
Let $V$ be an $n$-dimensional vector space over $\mathbb{C}$, and let $W$ be the $k$-fold tensor power $W=V\otimes \dots\otimes V$.
We have an obvious representation $\Psi$ of the symmetric group $S_k$ on $W$ given by $$ \Psi(\pi):v_1\otimes v_2\otimes\dots\otimes v_k \mapsto v_{1\pi}\otimes v_{2\pi}\otimes\dots\otimes v_{k\pi} $$ where linear maps and permutations are being written on the right.
The original question was the case $k=3$ and asked about simultaneous eigenvectors and eigenvalues of $S_3$.
In the more general case a natural question is (i) what is the multiplicity of the trivial representation in $\Psi$; (ii) what is the multiplicity of the sign representation in $\Psi$.
I believe that the answers can be expressed in terms of binomial coefficients; if so, is there an "easy" combinatorial explanation? (I say "easy", because the representation theory of the symmetric group is all combinatorics anyway - I am wondering if there's an immediate way to see what the answers to the questions are.)
Well as I suggested in the comments, Schur-Weyl duality gives a complete answer for how to decompose $V^{\otimes k}$ as an $S_k$ representation.
If you just want to understand the space of invariants, there is a purely combinatorial argument.
Let $N$ be an $n$ element set, and let's think of $V$ as $\mathbb{C} \langle N \rangle$ (the vector space of formal linear combinations of elements of $N$). The tensor power $V^{\otimes k}$ is then isomorphic to $\mathbb{C} \langle N^k \rangle$, with the action of $S_k$ coming from the action on $N^k$ by permuting the coordinates.
Here is a general lemma about combinatorial representation theory: If $G$ is a finite group acting on a finite set $X$, then the space of $G$-invariant vectors in $\mathbb{C}\langle X \rangle$ has a basis given by elements $\sum_{x\in O} x$ where $O$ is a G-orbit in $X$. In particular the dimension of the space of invariants is equal to the number of orbits.
Applying this to our setting where $G = S_k$ and $X = N^k$, we want to count the $S_k$ orbits on $N^k$. Two $k$-tuples in $N^k$ are in the same orbit iff for each $n \in N$ that $n$ appears the same number of times in the two $k$-tuples. Keeping track of the multiplicity of each $n \in N$, we can recognize this as a stars-and-bars type counting problem, which gives $\binom{n+k-1}{n-1} = \binom{n+k-1}{k}$ orbits -- and hence a that dimensional space of invariants.