Show $\phi : S_3 \rightarrow S_3$ given by $\phi(f^{i}g^{j}) = f^{2i}g^{j}$ is an automorphism
I'm not really sure how to begin.
I know that I need to show that $\phi$ is a group isomorphism, but the problem is more confusing for me when written in terms of $f$ and $g$.
For example, how do I show injective and surjective?
I am reframing the question a bit because the question is not well Defined,(And Op Should be more careful about Making and Suggesting questions)furthermore there will be ambiguity of understanding what $f,g$ actually is?. The question should be framed actually like this, Show that there exist, $f,g\in S_{3}$ such that there may exist automorphism $$\phi : S_3\longrightarrow S_3$$ satisfying $$\phi(f^{i}g^{j})=f^{2i}g^{j}$$
$\textbf{Proof:}$ We know that $S_3$ using generators can be represented as,
$$S_{3}=\{f,g,e,f^2,fg,f^{2}g|f^3=e , g^2=e , gf=f^{2}g\}$$ Now, we have $$\phi(f^{i}g^{j})=f^{2i}g^{j}$$ Note that this is similar to, $$\phi(f^{i\pmod 3}g^{j\pmod 2)})=f^{2(i\pmod 3)}g^{(j\pmod 2)}$$ Thus enough to consider $(i,j)\in\{0,1,2\}\times\{0,1\}=\{(0,0),(1,0),(2,0),(0,1),(1,1),(2,1)\}$ Trivial to check $$f^{0}g^{0}\longrightarrow e$$, $$f^{1}g^{0}\longrightarrow f^2$$,$$f^{2}g^{0}\longrightarrow f^4=f^{3}f=e.f=f$$, $$f^{0}g^{1}\longrightarrow g$$, $$f^{1}g^{1}\longrightarrow f^{2}g$$, $$f^{2}g^{1}\longrightarrow f^{4}g=fg$$
Hence, $\phi$ is a automorphism. $\blacksquare$
Remark/Try: Find all integers $(m,n)$ such that $$\phi_{m,n}(f^{i}g^{j})=f^{mi}g^{nj}$$ is an automorphism from $S_3$ to $S_3$ for some particular element $f,g\in S_3$ that may exist in $S_3$. Note each $(m,n)$ pair gives different functions.