Problem:
If $a, b \in G$ are generators of a group $G$, then there is an automorphism $f:G\to G$ such that $f(a)=b$.
Since $G=\langle a \rangle$ and $b\in G$, I proposed that $f$ be given by: $$ f(g)=g^t, \text{where } t\in\mathbb{Z} \text{ is the integer such that }a^t=b$$
This is an homomorphism, but I'm not sure if is injective and surjective since I have not been able to prove it so far.
Is this really an isomorphism? If it is, can you give me some hints to prove it please?
Edit#1: I already managed to solve the surjectivity part. Here is my attempt to prove injectivity.
Let $x \in \ker f$, this is $e=f(x)=x^t$
Since $G=\langle a \rangle$ and $x\in G$, there is a $k\in \mathbb Z$ such that $x=a^k$. Then, $e=x^t=(a^k)^t=(a^t)^k=b^k$.
Here I consider two different cases. If $G$ is infinite, since $G=\langle b \rangle$ all powers of $b$ must be different, therefore given that $e=b^k$ we have $k=0$. Consequently $x=a^k=a^0=e$.
If $G$ is finite, let's suppose $|G|=n$. Since $a$ and $b$ are generators, the order of both elements is $n$. From $b^k=e$ we have that $n$ divides $k$, this is, $k=nk'$. Therefore, $x=a^k=a^{nk'}=(a^n)^{k'}=e^{k'}=e$.
Is this correct? If so, do you think there is a more direct argument to prove it?
Here is a roadmap:
If $G=\langle a \rangle$ and $b \in G$, then there is an homomorphism $f:G\to G$ such that $f(a)=b$.
The image of $f$ is $\langle b \rangle$.
If $G=\langle b \rangle$, then $f$ is injective.
For the last step, you need to consider the two cases: $G$ finite and $G$ infinite.