Let $\mathbb{C}S_n$ be the group algebra of $S_n$ with basis $\{e_g\}_{g\in S_n}$. What can we say about the map $$(\mathbb{C}S_n)^{op}\to\text{End}(V^{\otimes n}),\quad e_g\mapsto (-).g,$$ where '$.$' means the right action of $S_n$on $V^{\otimes n}$?
It is for sure an algebra homomorphism. But is it injective or surjective?
$\def\End{{\rm End}} \def\sign{{\rm sign}} \def\id{{\rm id}} $
I assume $V$ is a complex vector space of dimension $r$. $\rm{GL}(V)$ also acts on $V^{\otimes n}$, say on the left, and the homomorphic image of $\mathbb{C}S_n$ is precisely the algebra of endomorphisms of $V^{\otimes n}$ which commute with the left action of $\rm{GL}(V)$. The homomorphism is injective if and only if $n \le r$. All this is known as Schur--Weyl duality. The algebra $\mathbb{C}S_n$ has minimal ideals labelled by Young diagrams of size $n$. The ideals which survive the homomorphism are those labelled by Young diagrams with no more than $r = \dim V$ rows (or columns in case you happen to have transposed the Young diagrams in the labelling of minimal ideals). You can read about this in many places, for example Goodman and Wallach, Symmetry, Representations and Invariants, Springer Verlag. (That's Roe Goodman, by the way, not me.)
Edit: I want to add some completely elementary observations that show that the map, $\varphi: (\mathbb C S_n)^{op} \to \End(V^{\otimes n})$ is never surjective and is "usually" not injective. The general analysis (e.g in Goodman-Wallach) is actually based on these observations.
First, if $T$ is in the image of $\varphi$, then $T$ has to satisfy the following condition. For $A \in \End(V)$, one can define $A^{\otimes n} \in \End(V^{\otimes n})$ so that $A^{\otimes n}(v_1 \otimes \cdots \otimes v_n) = Av_1 \otimes \cdots \otimes Av_n$. Any $T$ in the image of $\varphi$ has to commute with $A^{\otimes n}$ for all $A$. It is easy to invent operators in $\End(V^{\otimes n})$ which do not satisfy this condition, so $\varphi$ is never surjective.
Next for the failure of injectivity, take $V = \mathbb C^2$ with basis $\{e_1, e_2\}$. Let's show that $(\mathbb C S_3)^{op} \to \End(V^{\otimes 3})$ is not injective by finding something in the kernel. Consider $f \in \mathbb C S_3$ defined by $f = \sum_{\sigma \in S_3} \sign(\sigma) \sigma$. $V^{\otimes 3}$ has basis the set of simple tensors $e_{i_1} \otimes e_{i_2} \otimes e_{i_3}$ with the indices in $\{1,2\}$. Any such simple tensor has a repeated index. Let's show for example that $(e_1 \otimes e_1 \otimes e_2)\cdot f = 0$. The right coset representatives of $H = \{e, (1,2)\}$ are $\{e, (13), (23)\}$. Hence $f$ has the factorization $f = (e - (12)) (e - (13)-(23))$. It follows that $(e_1 \otimes e_1 \otimes e_2)\cdot f = 0$. By adapting this argument, show that $v \cdot f = 0$ for $v$ any basis vector $v = e_{i_1} \otimes e_{i_2} \otimes e_{i_3}$.