= Solution
A <permutation cycle> $(a_1\ a_2\ \cdots\ a_p)$ sends $a_1$ to $a_2$, each successive listed point to the next, and $a_p$ to $a_1$, fixing every unlisted point. Its length is $p$. For any <permutation> $h$,
$$
h(a_1\ \cdots\ a_p)h^{-1}=(h(a_1)\ \cdots\ h(a_p)).
$$
Thus <conjugate group elements> in the <symmetric group> have the same cycle length. Conversely, two cycles of the same length are conjugate: send their entries, in cyclic order, to one another and extend this <bijection> arbitrarily to the unused points. Length-one cycles represent the identity and cause no exception to this conclusion.
An odd-length <permutation cycle> is an <even permutation>, since its sign is $(-1)^{p-1}$. Let $c,d$ be such cycles on $p+2$ points and choose $h\in S_{p+2}$ with $hch^{-1}=d$. If $h$ is odd, let $t$ be the <transposition> of the two points unused by the listed cycle $c$. Then $tc=ct$, and $ht$ is even with
$$
(ht)c(ht)^{-1}=hch^{-1}=d.
$$
Hence \b[all the $p$-cycles are conjugate within $A_{p+2}$]. The two spare letters supply the parity correction; they are the reason the same argument need not work with only one spare letter.
In $A_4$ the answer is \b[no]. For example, $(1\ 2\ 3)$ and $(1\ 3\ 2)$ are conjugated by the odd <transposition> $(2\ 3)$ in $S_4$. The <centralizer> of $(1\ 2\ 3)$ in $S_4$ is exactly its order-three <cyclic subgroup>: a commuting <permutation> fixes the unique fixed point $4$ and acts as a power of the cycle on the other three points. All these centralizing elements are even. Every other conjugating element differs from $(2\ 3)$ by a centralizing element, so it is also odd. There is no conjugator in $A_4$. Equivalently, the eight three-cycles split into two <conjugacy classes> of size $12/3=4$, illustrating the <alternating conjugacy class splitting criterion>.
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