Let $E/k$ be an elliptic curve over a field of characteristic $\neq$ 2, 3. Then we have an isomorphism $ [ \ \ ] :\mu\_n \rightarrow\mathrm{Aut}\_{\overline{k}}(E)$, $[ \zeta ] : (x,y) \rightarrow (\zeta^2x, \zeta^3y) $, here $n=2, 4,6$, depending on the $j$-invariant $j(E) $. See Corollary 10.2 on Ch3 in "The arithmetic of Elliptic Curves" by Silverman. There it mentioned that this isomorphism commutes with the Galois action, but I am confused. For example, let $ \sigma \in G=\mathrm{Gal}(\overline{k}/k) $, then $[\zeta^\sigma] : (x,y) \rightarrow ( (\zeta^\sigma)^2x, (\zeta^\sigma)^3y)$, but $\sigma( [\zeta]) $ is $(x,y) \rightarrow (\zeta^2x, \zeta^3y) \rightarrow ( (\zeta^2x)^\sigma, (\zeta^3y)^\sigma)$, hence they are different. Am I thinking something in the wrong way? ( Sorry about such level of question....)



Automorphism groups of Elliptic curves as Galois module
It's as Matt indicates. You might find it easier to look at the commutative diagram, which is what it means for two maps to commute. Thus
$$
 \begin{aligned}
 E(\bar{k}) &\qquad\xrightarrow{[\zeta]} & E(\bar{k}) \cr
 \downarrow\sigma && \downarrow\sigma \cr
 E(\bar{k}) &\qquad\xrightarrow{[\zeta^\sigma]} & E(\bar{k}) \cr
 \end{aligned}
$$
The map $\sigma([\zeta])$ is the composition $\sigma\circ[\zeta]$, which from the diagram is equal to $[\zeta^\sigma]\circ\sigma$.


