Why does group action by conjugation on sylow subgroups define a homomorphism into the symmetric group? Sylow theorems state that sylow p subgroups of a group G are conjugate. Often I see argumentation that if there are n sylow p subgroups in G then we can define a group action on it by conjugation and hence create a homomorphism from G into Symmetric group or order n. Please provide a proof that this homomorphism is legitimate and why conjugation by any element on a sylow p subgroup takes you to another sylow p subgroup?
 A: 
"Please provide a proof that this homomorphism is legitimate".

Every action of a group $G$ on a set $S$ defines a homomorphism from $G$ to $\operatorname{Sym}(S)$ (by $g\mapsto(s\mapsto \varphi_g(s):=\mathcal{A}(g,s)$), and viceversa (by $(g,s)\mapsto \mathcal{A}(g,s):=\varphi_g(s)$), and the action of $G$ on $\operatorname{Syl}_p(G)$ by conjugation doesn't make an exception to this general fact about actions. Furthermore, given two equicardinal sets, say $X$ and $Y$, and a bijection $\alpha$ between them, it's easy to build up an isomorphism between $\operatorname{Sym}(X)$ and $\operatorname{Sym}(Y)$, and thence the above mentioned homomorphism can be thought of from $G$ to $S_{n_p}$ $($rather then to $\operatorname{Sym}(\operatorname{Syl}_p(G)))$, where $n_p:=|\operatorname{Syl}_p(G)|$.
A: In a group $G$ of cardinal $p^km$ with $gcd(p,m)=1$ a $p$-Sylow is defined to be any subgroup $S$ of $G$ whose cardinal is $p^k$. Suppose that $g\in G$ then $gSg^{-1}$ is a subgroup of $G$ and its cardinal must be the same as the cardinal of $S$ because, in a group $G$ conjugation by an element is a one-to-one morphism (it is also onto but you don't need this here), so it sends subgroups to subgroups and the cardinal is constant. Finally $gSg^{-1}$ is a subgroup of cardinal $|S|=p^k$ that $gSg^{-1}$ is also a $p$-Sylow. This answers your last question. 
For the first question. Suppose $X=\{S_1,...,S_n\}$ is the set of $p$-Sylows in $G$ (because of the first Sylow theorem, $n\geq 1$) with an arbitrary enumeration. It is easily seen by the remark above that $G$ acts on $X$ by conjugation. Namely if $S$ is a $p$-Sylow and $g\in G$ then :
$$g.S:=gSg^{-1}\in X$$
It is a group action because $1_G.S=S$ and $g.(h.S)=g(h.S)g^{-1}=ghSh^{-1}g^{-1}=(gh).S$. But you can see this action in a different way. Let $Bij(X)$ be the set of functions from $X$ to $X$ which are one to one and onto then define :
$$\rho: G \rightarrow Bij(X)$$
$$g\mapsto [S\mapsto gSg^{-1}] $$
The fact that $\rho(g)$ is one to one and onto comes from the fact that $\rho(g^{-1})=\rho(g)^{-1}$ furthermore from the rules of group action, $\rho$ is actually a group morphism from $G$  to $Bij(X)$ (with the law of composition).
Now the last thing to use is that $X$ is in bijection with $\{1,...,n\}$. Let us set :
$$\psi: \{1,...,n\}\rightarrow X $$
$$i\mapsto S_i $$
Then you have $\psi^{-1}\circ\rho(g)\circ\psi$ is a bijection of $\{1,...,n\}$ i.e. an element of $\mathfrak{S}_n$ (a permutation). So the morphism you are looking for is explicitely :
$$\rho_0:G\rightarrow \mathfrak{S}_n $$
$$g\mapsto \psi^{-1}\circ\rho(g)\circ\psi  $$
It should be remarked that this morphism is not canonical (it explicitely depends on your enumeration i.e. $\psi$) nevertheless, it is canonical up to conjugation in $\mathfrak{S}_n$. 
A: As elucidated in the other answers, we do have a group action of $G$ on the set of Sylow $p$-subgroups.  A group action always defines a homomorphism from the group into the group of symmetries of the set it's acting on (that is, into $S_n$, where $n$ is the number of
Sylow $p$_subgroups, in  this case).  Moreover, according to one of the Sylow theorems, all the Sylow subgroups are conjugate, so the action is in fact transitive.
