If $G$ acts on $V$, how does $G^n\rtimes S_n$ act on $V^{\otimes n}$? In a paper it was taken as obvious that if a finite group $G$ acts on a vector space $V$, then the semidirect product $G^n\rtimes S_n$ acts on $V^{\otimes n}$. I've tried to elaborate on how I think works below, I'm hoping someone can point out any mistakes, and or give a cleaner way to justify this than by long winded computation if possible.
Firstly I assume $S_n$ acts on $G_n$ in $G^n\rtimes S_n$ by permutations, i.e. $\sigma \rhd (g_1,\dots,g_n)=(g_{\sigma^{-1}(1)},\dots,g_{\sigma^{-1}(n)})$. The product in $G^n\rtimes S_n$ is $(\underline{g},\sigma)\cdot (\underline{h},\tau)=(\underline{g}\cdot (\sigma\rhd \underline{h}),\sigma \tau)$.
We have the following two obvious actions: $G^n$ on $V^{\otimes n}$ via $(g_1,\dots,g_n)\rhd v_1\otimes \dots\otimes v_n=(g_1\rhd v_1)\otimes \dots \otimes (g_n\rhd v_n)$, and $S_n$ on $V^{\otimes n}$ by permuting vectors, i.e. $\sigma \rhd v_1\otimes \dots \otimes v_n=v_{\sigma^{-1}(1)}\otimes \dots \otimes v_{\sigma^{-1}(n)}$. 
I then took the action of an element $((g_1,\dots,g_n),\sigma)\in G^n\rtimes S_n$ on $v_1\otimes \dots v_n$ as $(g_1 \rhd v_{\sigma^{-1}(1)})\otimes \dots \otimes (g_n\rhd v_{\sigma^{-1}(n)})$. This step didn't seem natural or obvious to me. Using this action, and the product on $G^n\rtimes S_n$ as above, we find $(\underline{1},1)\rhd v_1\otimes \dots \otimes v_n=v_1\otimes \dots \otimes v_n$, and also : $(\underline{g},\sigma)\cdot (\underline{h},\tau)\rhd v_1\otimes \dots v_n=(\underline{g},\sigma)\rhd [(\underline{h},\tau)\rhd v_1\otimes \dots \otimes v_n]$, as required for this to be an action.
 A: Maybe what you're missing is an understanding of the universal property of the semi-direct product. 
Indeed, a representation of $G^n\rtimes S_n$ on $V^{\otimes n}$ is nothing but a morphism $G^n\rtimes S_n\to GL(V^{\otimes n})$, and you already have obvious morphisms $G^n\to GL(V^{\otimes n})$ and $S_n \to GL(V^{\otimes n})$ and the "annoying computation" you do at the end is to prove a sort of compatibility between those two. 
To understand why that computation is natural and how you can think of it, it's interesting to understand what maps out of a semi-direct product do, what they look like. 
As a first approximation, we have the following : 

In the category of abelian groups, the direct product is a coproduct.

If you don't know that categorical jargon, it's ok: what this means is just the very intuitive idea that in the abelian world, if you have groups $A,B,C$ and maps $f: A\to C, g: B\to C$, you automatically get a map $A\times B\to C$, simply defined by $(a,b)\mapsto f(a) + g(b)$. 
This works well because everything is abelian. Well in fact you don't need that full hypothesis, and so a subtler result (which is just as easy to prove) is : 

Let $G,H,Q$ be groups. Then a map $G\times H\to Q$ is entirely determined by the maps $G\to Q$ and $H\to Q$, as long as their images commute.

That is, as long as $f(g)$ and $f'(h)$ commute for all $g\in G, h\in H$, you can do the same thing as in the abelian world and define $(g,h)\mapsto f(g)f'(h)$. Of course, this is a necessary condition because $(g,1)$ and $(1,h)$ commute in $G\times H$. 
Ok so that's a first approximation, and I wrote it to understand where the next step comes from - so that it's at least a bit motivated. 
Indeed, the difference between direct product and semi-direct product is that in the former you have a "clean" commutation : $gh = hg$, whereas in the latter there's another formula : $hg = (h\cdot g)h$ where $H$ acts on $G$ (and I let $h\cdot g$ denote this action)
Well in the exact same way as the above proofs, we get the following "universal property" : 

Suppose $H$ acts on $G$ by automorphisms, and let $Q$ be another group. Then a map $F: G\rtimes H\to Q$ is entirely determined by maps $f: G\to Q$ and $f': H\to Q$, as long as they satisfy $f'(h)f(g) = f(h\cdot g)f'(h)$. 

That is, to define a map out of a semi-direct product, you only have to define it on the two factors, and check this "commutation" hypothesis. 
Just as before, it's pretty easy to prove, but it's a good guiding principle. Especially if you rewrite it in terms of conjugation : $f'(h)f(g)f'(h)^{-1} = f(h\cdot g)$
In your example, you already have maps $G^n\to GL(V^{\otimes n}), S_n\to GL(V^{\otimes n})$, so these become maps on the semi-direct product if and only if you have the correct conjugation relation in $GL(V^{\otimes n})$ - that is probably what your computations looked like (although yours were maybe a bit more complicated, because you had a $g$ an $h$, a $\sigma$ and a $\tau$, whereas you could reduce to only a $g$ and a $\sigma$). 
The definition of the action you obtained is guided by the idea that $(g,1)(1,\sigma) = (g,\sigma)$, and this determines the action/the morphism entirely - that's where the step that didn't seem natural or obvious to you comes from. 
