If $G$ is a connected Lie group acting on a vector $\mathbb{C}$-space $V$ then it is well known that the algebra of invariants $\mathbb{C}[V]^G$ coincides with the algebra of invariants $\mathbb{C}[V]^L$ of corresponding Lie algebra $L.$

**Question.** Let now $G$ be a finite group, $V$ be its representation. Is there exists a Lie algebra $L$ and its representation on $V$ such that $\mathbb{C}[V]^G=\mathbb{C}[V]^L$?

It is easy to see that it impossible for symmetric group $S_n$. But maybe there are classes of finite groups for which can be found the positive answer?

**Edit.** Let $V=< v_1,v_2,\ldots,v_n >$ be standard representation of the symmetric group $S_n.$ Suppose that there exist a derivation $D=f_1 \frac{\partial}{\partial x_1}+\cdots+ f_n \frac{\partial}{\partial x_n}, f_i \in S(V)$ of the symmetric algebra $S(V)$ such that $D(I)=0$ for all $I \in \mathbb{C}[V]^{S_n}$.
Let $I_1,I_2,\ldots, I_n \in S(V)$ be a minimal generating set for $\mathbb{C}[V]^{S_n}.$ Since $D(I_k)=0, \forall k,$ we get the system of polynomial equations on $f_1,f_2,\ldots,f_n:$
$$
f_1 \frac{\partial I_1}{\partial x_1}+\cdots+ f_n \frac{\partial I_1}{\partial x_n}=0,\\
f_1 \frac{\partial I_2}{\partial x_1}+\cdots+ f_n \frac{\partial I_2}{\partial x_n}=0,\\
\ldots \\
f_1 \frac{\partial I_n}{\partial x_1}+\cdots+ f_n \frac{\partial I_n}{\partial x_n}=0.
$$
The Jacobian of system of invariants $I_1,I_2,\ldots, I_n$ is not zero. It follows the system has only trivial solution $f_i=0.$