What does $k[x_1,\dots,x_n]^G＝k$ mean for subgroup $G\subset k^*$？

Question

Let $k$ be a field (if necessary algebraically closed) and $G$ be a subgroup of multiplication group $k^＊$.

Now since $k^＊$ act on a ring $k[x_1,\dots,x_n]$, $G$ act.

If $k[x_1,\dots,x_n]^G＝k$, do we get $G＝k^＊$？　If not, what we know about $G$？

Answer 1

If you meant that $k^*$ acts through $g.f(x) = f(gx)$ then

$k[x_1,\dots,x_n]^G＝k$ iff $G$ is not a finite subgroup of $k^*$.

($g.\sum_\alpha c_\alpha x^\alpha=\sum_\alpha c_\alpha x^\alpha$ means that for each $c_\alpha\ne 0,g.x^\alpha= x^\alpha$ so that $g^{|\alpha|}=1$)