I am preparing for an exam in elementary group theory and while studying a proof by my professor, I came across something I just don't seem to understand. In his proof he claims that, considering a map from $(Z/2Z)^n$ to itself ($Z$ being the set of integers), the correspondence $(x_1 , ... , x_n)\to(x_2 , ... , x_n , x_1)$ is an isomorphism. Why is this correspondence an isomorphism?
Also, is it true that the correspondence $(x_1 , ... , x_n)\to(x_{a_1}, ... , x_{a_{n-1}} , x_{a_n})$, where $0 < a_i < n+1$ and $a_i = a_j <=> i=j$, is also an isomorphism from $(Z/2Z)^n$ to itself?
Hints
Fill in details :
Observe that $\;G:=\left(\Bbb Z/2\Bbb Z\right)^n\;$ can be seen as a vector space over the field $\;\Bbb F_2:=\Bbb Z/2\Bbb Z\;$, with the obvious termwise addition and scalar product.
Then, an automorphism of the abelian group $\;G\;$ is the same as an automorphism of the vector space $\;G_{\Bbb F_2}\;$ , and the map you show there is the one determined by the linear operator changing any ordered basis $\;\{e_1,...,e_n\}\;$ into the ordered basis $\;\{e_2,...,e_n,e_1\}\;$