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Let $G$ be a finite group with $n$ elements acting on $k[x_1,\cdots,x_n]$ through the regular representation. Is it true that $G$ is a subgroup (in the sense that there exists an injective group homomorphism) of the automorphism group $Aut(V)$ of the variety $V:=k^n/G$? If so, where can I read a proof for this? Thanks for your help!

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    Are you envisaging some sort of natural map? The automorphism group of the variety is typically huge (in the examples I computed) and might just include a copy of the group for no particular reason. This question seems very pathological unless there's some natural map from the group to the automorphisms which I've missed. – 2017-01-29
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    I don't know if there exists a natural map. That's why I am asking. I did some computer experiments with SAGE for the group $C_3$ and the Klein four group $C_2 \times C_2$ and there I could find an action of the given group to the variety, and this made me think if this is always the case ( for every finite group) . However I must admit, that the found action is far from being natural, as far as I can see. – 2017-01-29
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    The question (and the answer you accepted) is unclear. Do you mean whether $G$ "is isomorphic" to a subgroup of automorphisms of the quotient of the affine space $k^n$ by $G$? or do you mean in some prescribed way (related to the original action)? Since the answer includes an additional $X$, I don't see how it's related to the question. – 2017-01-29
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    Yes, I mean isomorphic to a subgroup. I think X is $k^n/G$. – 2017-01-29
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    The $X$ in the answer can't be what you suggest, as if it's finite Galois over a curve then it's a curve. The question is very pathological and the answer you accepted seems to give little indication as to how it's related to the question -- I think this is probably why you're picking up downvotes. The question is also a little imprecise -- you don't make it clear in which category you're computing the automorphism group. Is the quotient an algebraic variety? If so you'd be better off talking about affine $n$-space rather than $k^n$. – 2017-01-30
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    Thanks for your comment. I must admit that I do not understand everything written in the answer. I accepted, because the answerer seemed to know what he was talking about. I also asked, where I could read this up in a more detailed way, so that I understand the proof. Until noone is giving a better answer I will leave it as it is. – 2017-01-30
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    @stackExchangeUser To understand my answer perhaps you should start by reading and understanding the chapter on covering spaces in Hatcher's "Algebraic Topology". From there, it suffices to know that the theory of topological covering spaces of Riemann surfaces (with finitely many punctures) is equivalent to the theory of finite etale covers of the corresponding (finitely punctured) complex algebraic curves. On either side, you can always fill in the punctures to form a "complete curve", in a way that respects the Galois action. – 2017-01-30
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    @oceimon: Thanks again for your comments and answer! – 2017-01-30
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    @stackExchangeUser But, my answer only answered my interpretation of the question in the post title. I don't really understand the question in your actual post. Why would you expect $G$ to act on $k^n/G$? Can you explain the motivation for your question? – 2017-01-30
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    @oxeimon: The motivation was that I found for small groups ($C_3$ and $C_2 \times C_2$ ) some action from $G$ to $k^n / G$ and I wondered if this was a coincidence or if this is always the case (for every finite group). – 2017-01-30
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    For cyclic groups $G$ this is obvious —over the complex numbers at least. The multiplicative group of the field acts by multiplication on the variables on $k[x_1,\dots,x_n]$ and that action restricts to a faithful action on the invariant subring (as $\sum x_i$ is in that subring) Every finite subgroup of the multiplicative group therefore acts faithfully on the quotient in some way and, in particular, so does $G$. – 2017-01-30
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    If $A$ is the polynomial ring generated by variables $x_g$ with $g\in G$, then there is an action of $G\times G^{op}$ on $G$, with $(g,h)\cdot x_k=x_{gkh^{-1}}$. Since $G$ is normal in the product, this gives an action of $G^{op}=(G\times G^{op})/G$ on $A^G$. Is it faithful? As $G$ and $G^{op}$ are isomorphic groups, this would answer the question in general. It looks doubtful, though. – 2017-01-30
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    @MarianoSuárez-Álvarez: Thanks for your comment. Do you mean there is an action from $G \times G^{op}$ on $A$? (You wrote on $G$). What is $G^{op}$? Would you write your comment as an answer? – 2017-01-30
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    The opposite group. Indeed, the group acts on itself by left multiplication and by right multiplication, and the two actions commute, so that you get an action of the direct product. – 2017-01-30

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The answer to the question in your title is yes. This follows from the fact that every finite group $G$ is a Galois group of a finite Galois covering $X\rightarrow\mathbb{P}^1_{\mathbb{C}}$, and hence $G\subset\text{Aut}_{\mathbb{C}}(X)$.

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    Thanks for your answer. Do you know where I can read this in a more detailed way? – 2017-01-29
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    @stackExchangeUser The etale fundamental group of $P^1_{\mathbb{C}}$ minus $n+1$ points is the free group on $n$ generators. Its quotients $G$ correspond to finite $G$-Galois covers of $\mathbb{P}^1$ ramified only over those $n+1$ points. Thus, every finite group $G$, generated by $n$ elements, is the Galois group of some cover of $\mathbb{P}^1_{\mathbb{C}}$ ramified over $n+1$ points. – 2017-01-29