On the field of invariants of a finite group So let $G$ be a finite group and let $\iota:G\rightarrow S_n$ be an embedding of $G$ in a symmetric group of degree $n$ for some fixed integer $n$. Let $K$ be a fixed field of characteristic $0$. The group $S_n$ permutes the variables $\{x_1,\ldots,x_n\}$ and therefore
acts on the field 
$$
L:=K(x_1,\ldots,x_n).
$$
One may look at the invariant subfield $L^{G}\subseteq L$. From Galois theory one has that
$L/L^G$ is a Galois extension with Galois group $G$. In particular, the transcendence degree
of $L^G$ over $K$ is equal to $n$. In general, the field $L^G$ is not purely transcendental
so the following question makes sense:
Q: Does the isomorphism class of $L^G$ depend on the embedding $\iota$ ?
Intuitively I would say no, but this is really just a guess!
 A: It seems likely (and may be known) that it does depend on the permutation representation. However, it is a subtle question as the stable isomorphism class (of any faithful permutation, or in fact arbitrary, representation) does not. Here two extensions $K$ and $K'$ of a field $k$ are stably isomorphic if for some $m$ and $n$ $K(x_1,\ldots,x_m)$ and $K'(x'_1,\ldots,x'_n)$ are isomorphic as $k$-extensions.
Addendum: The result is well-known but I cannot at the moment come up with a reference so instead I give the proof: Let $V$ be a faithful $G$-representation and $U$ the non-empty Zariski open subset where $G$ acts freely. Then $k(V)^G$ is the fraction field of $U/G$. If now, $V'$ is another faithful representation with open subset $U'$ we have that $U\times V'$ has a free $G$-action with a linear action on the second factor. Hence $U\times V'/G$ is a vector bundle over $U/G$ (by descent theory) and in particular its fraction field is stably isomorphic to that of $U/G$. However, $U\times V'/G$ is birational to $U\times U'/G$ which in turn is birational to $V\times U'/G$. The fraction field of the latter is for the same reason stably isomorphic to that of $U'/G$. 
A: If I were going to think about this I would do the following.  Take some situation where I know Noether's problem has a negative answer, like the generalized quaternion group of order 16 in its regular permutation representation on 16 variables.  (I didn't know that off the top of my head, and maybe it's not the easiest example -- it came up on Google, ascribed to Serre.)
Now this group has lots of embeddings in S_16, I suppose.  There are so many cases where Noether's theorem is known to have a positive answer that you might be able to show that one of these has a rational field of invariants. 
