Suppose $G$ is a group and $A$ is a finitely generated algebra over a field $\mathbb{k}$. Let $X=\{x_1,...,x_n\}$ be a set of generators for $A$, and suppose $G$ acts on $X$. Is this enough to define a group action by $G$ on all of $A$?
My thinking was as follows: Since $X$ generates $A$ as an algebra, any element $x \in A$ may be written as
$x= \sum_i c_i x_1^{a_{1_i}}...x_n^{a_{n_i}}$
where $c_i \in \mathbb{k}$ for all $i$. If $g \in G$, we define
$g \cdot x := \sum_i c_i (g \cdot x_1)^{a_{1_i}}...(g \cdot x_n)^{a_{n_i}}.$
Does this give us a well defined group action on $A$? This works in the case of polynomial rings, though I'm not sure if it works in the case of more general finitely generated algebras, say a quotient of a polynomial ring.
If your generators are algebraically independent over $k$. Then it is enough, because construction of polynomial algebra gives rise to a functor: $$F:\mathbb{Set}\rightarrow \mathbb{Alg}_k$$ So if you have a group $G$ acting on a set by bijections $\{\phi_g\}_{g\in G}$, then you have action of $G$ via elements $\{F(\phi_g)\}_{g\in G}$ on $F(X)=k[X]$.
On the other hand if you have some relations between generators, then it might not be true. Take for example: $$A=k[\overline{x},\overline{y}]=k[x,y]/(y^2)$$ and action of $\mathbb{Z}_2$ on $\{\overline{x},\overline{y}\}$ by $\overline{x}\mapsto \overline{y}$. There is no $k$-automorphism of $A$ sending $\overline{x}\mapsto \overline{y}$, because $\overline{y}^2=0$ and $\overline{x}^2\neq 0$.