I'm following the discussion in Artin's Algebra. Artin defines a permutation representation of a group $G$ as a homomorphism from the group to the symmetric group:$$\varphi:G\rightarrow S_n$$
Then there's a proposition: There is a bijective correspondence between operations of $G$ on the set $S = \{1,..., n\}$ and permutation representations $\varphi:G\rightarrow S_n$ $$\Bigg[\textrm{ operations of G on S }\Bigg]\longleftrightarrow \Bigg[\textrm{ permutation representations }\Bigg] $$
Proof: Define a permutation representation $\varphi$ by setting $\varphi(g) = m_g$, left multiplication by $g$. Conversely, the same formula defines an operation of $G$ on $S$.
Question 1: Why does he only consider the operation "multiplication by $g$"? There are other operations like conjugation by $g$. Shouldn't these other operations also have corresponding perm. reps.?
Question 2: How about the operations of $S_3$ on the set of 2 elements $S = \{a, b\}$? If we are defining each homomorphism $\varphi(g) = m_g$ by left multiplying the set $S$ by $g \in S_3$, how does e.g. $(123) \in S_3$ act on $b\in S $?
I think you have some things confused here.
The left side is the set of operations (also called actions) of $G$ on $S$, which are maps $G \times S \to S$ satisfying the properties of a group action. Given a $g \in G$, "left multiplication by $g$" means the map $s \mapsto gs$. Calling it "left multiplication" can be a bit misleading because there is no corresponding "right multiplication" (nor "conjugation" as you talk about in Q1), and it's not a "multiplication" in the sense of the multiplication in $G$. Remember that $G$ is acting on $S$, not on itself.
In Q2, your question is phrased a bit strangely because it sounds like you may be thinking there's only one way for $S_3$ to act on $\{a,b\}$. But that's not the case. $S_3$ can act on $\{a,b\}$ in any way that satisfies the properties of a group action. And $m_g$ is not defined until you have chosen an action of $G$ on $S$. Once you have chosen an action, then $m_g$ is the map sending $s \mapsto gs$; but $gs$ doesn't make sense unless you have chosen an action. In fact, there are two ways for $S_3$ to act on $\{a,b\}$.