= Solution
{c}
For $n\ge2$ with $n\ne6$, bijectivity of $\phi$ and the previous part imply
$$
\frac{n!}{2^k k!(n-2k)!}=|T_k|=|T_1|=\frac{n(n-1)}2.
$$
The supplied numerical fact forces $k=1$. Thus $\phi$ permutes the <transpositions>.
Suppose $n\ge3$, and write $s_r=\phi((1r))$ for $r=2,\ldots,n$. These are distinct <transpositions>. For $r\ne t$, the product $(1r)(1t)$ has <order of a group element> three, so $s_rs_t$ also has order three. Two distinct <transpositions> have this property precisely when their two-element sets of moved points meet in one point: disjoint ones have a product of order two, whereas ones sharing one point have a three-cycle as product. The two-element <support of a permutation> associated with each $s_r$ therefore meets that of every other $s_t$.
We now show that they have a common point. Choose two; the <support of a permutation> is $\{a,b\}$ for one and $\{a,c\}$ for the other, where $a,b,c$ are distinct. Any further two-element set meeting both either contains $a$ or is $\{b,c\}$. If $\{b,c\}$ occurs, any two-element set meeting all three must itself be one of $\{a,b\},\{a,c\},\{b,c\}$: a set containing $a$ must also contain $b$ or $c$, and a set omitting $a$ must contain both $b,c$. Hence a pairwise-intersecting family with no common point consists of exactly those three edges of a triangle.
For $n\ge5$, there are $n-1\ge4$ distinct $s_r$, ruling out that triangle. For $n=3$, the two sets already have a common point. For $n=4$, the triangle alternative would make all three $s_r$ fix the fourth point. They would then generate a proper <subgroup> of $S_4$, contrary to surjectivity of $\phi$ and part (a)(iv), which imply that the $s_r$ generate $S_4$. Thus in every case $n\ge3$ there is a common point $a$, and we may write
$$
s_r=(a\ b_r)\qquad(2\le r\le n).
$$
The $b_r$ are distinct and different from $a$, so they exhaust the remaining $n-1$ points. Define a <permutation> $\sigma$ by
$$
\sigma(a)=1,\qquad \sigma(b_r)=r.
$$
Then
$$
\sigma^{-1}(1r)\sigma
=(\sigma^{-1}(1)\ \sigma^{-1}(r))
=(a\ b_r)=\phi((1r)).
$$
The two <group homomorphisms> $\tau\mapsto\phi(\tau)$ and $\tau\mapsto\sigma^{-1}\tau\sigma$ agree on a <generating set of a group>, so agree on every product of its elements, hence throughout $S_n$:
$$
\boxed{\phi(\tau)=\sigma^{-1}\tau\sigma\qquad(\tau\in S_n).}
$$
Finally, $S_1$ is trivial and $S_2$ has a unique nonidentity element, so every <automorphism> of either is the identity and we can take $\sigma=e$. This completes all degrees $n\ne6$ and proves that every such <automorphism> is an <inner automorphism>.
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