I am trying to understand a proof that $S_3$ is isomorphic to $\operatorname{Aut}(S_3)$ located on page 2 of this link: http://www.math.ntu.edu.tw/~hchu/Algebra/Jacobson%5B19%5D.pdf
I'm going to rewrite the proof in my own words and pause at the steps that I can't follow.
We have $$S_3 = \{e, a = (123), a^2 = (132), b = (12), ab = (123)\},$$ so, e.g., $S_3 = \langle (123), (12) \rangle$. It follows then that $ba = a^2 b$, which implies that \begin{align*} (a^m b^n)(a^p b^q) = a^{m+(n+1)p} b^{n+q}, \end{align*} where $m,p \in \{0,1,2\}$ and $n,q \in \{0,1\}$.
My first point of confusion is trying to verify this. One possibility is running over cases, but there must be an easier way. First, the claim seems to be that because $S_3$ is generated by $a$ and $b$, I can write any element $x \in S_3$ as $x = a^m b^n$ where $m \in \{0,1,2\}$ (since the order of $a$ is $3$) and $n \in \{0,1\}$ (since $b$ has order $2$). The problem then becomes that I don't know how to introduce $ba = a^2 b$ since I don't know if I have two factors of $a$ to work with. I assume this boils down to an associativity argument, but I'm struggling to see how to do it.
Given any automorphism $\phi$ acting on $x \in G$, $\phi(x)$ has the same order as $x$. Then $\phi(a) = a^i$ and $\phi(b) = a^j b$ where $i = 1,2$ and $j = 0,1,2$.
I understand the first sentence but not what follows. $a$ has order $3$, but upon mapping it to $a^I$, it no longer has order $3$; similarly for mapping $b$, an element of order $2$, to $a^j b$. Am I missing something here?
The remainder of the proof follows from these facts, so I think if I could understand these steps, I could probably make sense of the rest.