If $G$ is a cyclic group where both $a$ and $b$ are generators. How would I prove that $f:G \rightarrow G$ given by $f(a^i)=b^i$ is an automorphism.
I know that an automorphism is the identity map. So if $G$ is cyclic of order $n$. So I believe I have to show that $f$ is a well defined homomorphism, but I'm not exactly sure how to do this.
To show that the map is well-defined, show that if $a^i = a^j$, then $b^i = b^j$. Remember that in any group $G$, $a^i = a^j$ if and only if $i$ and $j$ are congruent modulo the order of $a$ in $G$.
The rest is routine, you just need to remember that since $a$ and $b$ are generators, any element of $G$ can be written as a power of $a$ and as a power of $b$ (in general with different exponents).