In this exercise we shall prove that $Aut(S_n) =S_n$ for $n > 6$. (The results holds true for $n = 4, 5$ too and fails for $n = 6$.) Thus, $S_n$ is complete for $n > 6$.
(a) Prove that an automorphism of $S_n$ takes an element of order 2 to an element of order 2.
(b) For $n > 6$ use an argument involving centralizers to show that an automorphism of $S_n$ takes a transposition to a transposition.
(c) Prove that every automorphism has the effect $(12)\to (a b_2), (13) \to (a b_3),\ldots, (1_n) \to (a b_n)$,for some distinct $a, b_2,\ldots,b_n \in\left\{1, 2,\ldots, n\right\}$. Conclude that $|Aut(Sn)|\leq n!$.
(d) Show that for $n > 6$ there is an isomorphism $S_n\simeq Aut(S_n)$.In this exercise we shall prove that $Aut(S_n) = S_n$ for $n > 6$.
My solution:
(a) Let $\phi:S_n\to S_n$ automorphism. Let $\sigma \in S_n$ of order 2. Then $(\phi(\sigma))^2=\phi(\sigma^2)=\phi(e)=e$.
(b) (I have not idea :( ) Only I have that, if $\sigma\in Aut(S_n)$ and $K$ conjugaccy class, then $\sigma(K)$ is a conjugaccy class.
(c) (Proof Idea) Let $(1r), (1s) \in S_n,\ r\neq s,\ r,s\neq 1$ and $f\in Aut(S_n)$. By (b), $f(1r)=(ab_2)$ and $f(1s)=(a'b_3)$. Then $f(rs)=f((1r)(1s)(1r))=(a'b_3)(ab_2)(a'b_3)$ but $f(rs)$ is a transposition. If $(a'b_3), (ab_2)$ are disjoints then $f(rs)=(ab_2)=f(1r)$ a contradiccion because $f$ is injective. Therefore $(a'b_3),(ab_2)$ are not disjoints. Suppose $a=a'$.
Now, let $(1t)$ other transposition $(t\neq 1, t\neq r,s)$. Then $f(1t)=(a''b_4)$. WIth the same argument, $f(1t)=(b_2b_4)$ etc... Finally, let $a,b\in \left\{2,\ldots,n\right\}$ with $a\neq b$ then $(ab)=(1a)(1b)(1a)$ then $\left\{(1a):a=2,\ldots,n\right\}$ generates any transposition, so generate $S_n$. Therefore, any $f\in Aut(S_n)$ is determined by $a,b_2,b_2,\ldots, b_n$. Because $a$ has $n$ possible value, $b,\ n-1$ possible value, $\ldots$, then exists at most $n!$ automorphisms. Therefore $|Aut(S_n)|\leq n!$.
(d) Because in general, $\phi:G\to Aut(G): g\mapsto \tau_{g}$ with $\tau_g(x)=gxg^{-1}$ is a homomorphism. Let $\phi:S_n\to Aut(S_n)$ homomorphism. By First Isomorphism Theorem, ${S_N}/{ker(\phi)}\simeq \phi(S_n)$ and $ker(\phi)=Z(S_n)=\left\{(1)\right\}$ then $\phi(S_n)\simeq S_n$ then $|\phi(S_n)|=n!$ and by (c) $|Aut(S_n)|\leq n!$, then $Aut(S_n)\simeq S_n$.
How could (b) be probed with a centralizers argument? I don't quite understand how to go.
Actualization 1. With help from Arturo Magidin, I have the following:
Affirmation 1. For any $(ab),(c,d)\in S_n$ with $a\neq b, b\neq d |C_{S_n}(ab)|=|C_{S_n}(cd)|$.
Proof: $(ab)=\sigma(ab)\sigma^{-1}=(\sigma(a)\sigma(b)) and (cd)=\tau(cd)\tau^{-1}=(\tau(c)\tau(d))$ then $C_{S_n}(ab)=\left\{\sigma\in S_n:\sigma(a),\sigma(b)\in\left\{a,b\right\} \right\}$ and $C_{S_n}(cd)=\left\{\tau\in S_n:\tau(c),\tau(d)\in\left\{c,d\right\}\right\}$ then the order are equal. Now, if $f(ab)$ is not a transposition then $f(ab)=(a_1b_1)\cdots (a_kb_k)$ with $a_i\neq b_i$, disjoint descomposition. Now, $|C_{S_n}(ab)|=|C_{S_n}(f(ab))|$ because $f$ is a automorphism.
I want proves that $|C_{S_n}(ab)|=|C_{S_n}(f(ab))|$ is a contradiction. I have the following idea:
$\sigma \in C_{S_n}(f(ab))$ then $\sigma(f(ab))\sigma^{-1}=(ab)$ then $\sigma(a_1b_1)\sigma^{-1}\cdots \sigma(a_kb_k)\sigma^{-1}=(\sigma(a_1)\sigma(b_1))\cdots (\sigma(a_k)\sigma(b_k))$.
by affirmation 1, $|C_{S_n}(ab)|=|C_{S_n}(a_ib_i)|$ and from the two previous facts I should prove that $|C_{S_n}(ab)|\neq |C_{S_n}(f(ab))|$ but I don't know how to formalize this idea.
Actualization 2. $|C_{S_n}(ab)|=2(n-2)!$.
proof. Let $\sigma\in C_{S_n}(ab)$ then $\sigma(ab)\sigma^{-1}=(ab)$ then $(\sigma(a)\sigma(b))=(ab)$ then $\sigma(a)$ it can take two possible values. and plus there are $(n-2)!$ ways to permutates $\left\{1,\ldots,n\right\}\setminus\left\{a,b\right\}$. Therefore $|C_{S_n}(ab)|=2(n-2)!$
Affirmation 2:
$|C_{S_n}(f(ab))|=(2k)(2k-2)(2k-4)\cdots (2)[(n-2k)!]$.
Proof. Let $\sigma\in C_{S_n}f(ab)$ then $\sigma f(ab)\sigma^{-1}=f(ab)$ then $(\sigma(a_1)\sigma(b_1))\cdots (\sigma(a_k)\sigma(b_k))=(a_1b_1)\cdots (a_kb_k).$ Now, $\sigma(a_1)$ can take $2k$ possible value $(a_1,b_1,\cdots, a_k,b_k)$.
$\sigma(a_2)$ can take $2k-2$ possible value.
$\vdots$
$\sigma(a_k)$ can take $2$ possible value
Therefore, exists $(2k)(2k-2)(2k-4)\cdots (2)$ possible value for $\sigma(a_1),\sigma(b_1),\cdots, \sigma(a_k),(\sigma(b_k))$ and plus there $(n-2k)!$ way to permutates $\left\{1,\ldots,n\right\}\setminus\left\{a_1,b_1,\cdots, a_k,b_k\right\}$.
It is correct?
Your answer for (a) is incomplete. You have shown that an element of order $2$ has image of exponent $2$, but you have not proven that it is of order $2$. You need to justify that the order is exactly $2$ and not merely a divisor of $2$.
(b) Elements of order $2$ are products of transpositions. Note that $x$ centralizes $\sigma$ if and only if $\phi(x)$ centralizes $\phi(\sigma)$. So if you can show that you can tell that an element of order $2$ is a single transposition by the size of its centralizer, then you can use that to show that the image of a transposition must be a transposition. For instance, compare the size of the centralizer of $(12)$ with the size of the centralizer of $(12)(34)$...