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 and $\mathbb{C}[V]^G$ be the corresponding  algebra of invariants. 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 $D=f_1 \frac{\partial}{\partial x_1}+\cdots+ f_n \frac{\partial}{\partial x_n}$  of the symmetric algebra $S(V)$ such that $D(I)=0$ for all $I \in \mathbb{C}[V]^{S_n}$.
Let $I_1,I_2,\ldots, I_n$ 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 $f_i=0.$