= Solution
A <rim hook> is a connected skew diagram containing no $2\times2$ square, whose removal leaves a <partition of an integer> diagram. Its leg length is its number of occupied rows minus one. The <Murnaghan–Nakayama rule> states that, if a <permutation> has cycle type $(k,\rho)$, then
$$
\boxed{\chi^\lambda(k,\rho)=
\sum_{\eta:\lambda/\eta\text{ a rim hook of size }k}
(-1)^{\operatorname{leg}(\lambda/\eta)}\chi^\eta(\rho).}
$$
Iteration removes one hook for each cycle length; the empty diagram has <character> one at the empty cycle type.
For shape $(3,3,3)$ and cycle type $(4,3,2)$, there are exactly two removable four-cell <rim hooks>. The remaining shapes are $(3,2)$ and $(2,2,1)$; the removed hooks have leg lengths one and two respectively. In shape $(3,2)$ the only removable three-cell <rim hook> leaves $(1,1)$ and has leg length one. In shape $(2,2,1)$ the only such hook leaves $(2)$ and again has leg length one. Since the <characters> of $(1,1)$ and $(2)$ at a transposition are $-1$ and $1$, we obtain
$$
\chi^{(3,2)}(3,2)=1,\qquad
\chi^{(2,2,1)}(3,2)=-1,
$$
and hence
$$
\boxed{\chi^{(3^3)}((1234)(56)(789))=-1-1=-2.}
$$
The cycle lengths can be taken in any order; using $4,3,2$ makes the two branches of the computation explicit.
Now consider the ordinary induction product $[\alpha][\beta]$. A <hook partition> is $(x,1^y)$, equivalently a <partition of an integer> whose second row has length at most one. If a hook shape $\nu$ occurs in the <Littlewood–Richardson rule>, then $\alpha\subseteq\nu$, forcing $\alpha$ itself to be a hook. In the skew hook $\nu/\alpha$, the new cells consist of a segment in the first row and a segment down the first column, with no shared column between those two segments. Every first-row entry must be $1$: its rightmost entry is the first letter of the <lattice word>, hence is $1$, and weak row increase forces all entries to its left to be $1$ too. The column segment has strictly increasing entries. Thus only $1$ can be repeated, forcing the content $\beta$ to be a hook as well. This also proves the assertion when the top segment is empty, since then every entry lies in a strictly increasing column.
Write the two nonempty hook factors as $\alpha=(a,1^k)$ and $\beta=(b,1^l)$, with $k=n-r-a$ and $l=r-b$. Let the prospective resulting hook have top-row length $x$ and leg length $y$. Its new top segment has $H=x-a$ cells and its new column segment has $V=y-k$ cells. The content consists of $b$ ones and the single letters $2,\ldots,l+1$. All those larger letters must be in the new column; its strict increase allows either no $1$ or exactly one $1$. There are therefore only two possibilities:
$$
(H,V)=(b,l)\quad\text{or}\quad(b-1,l+1).
$$
In the first case fill the top segment with $b$ ones and the column with $2,3,\ldots,l+1$. In the second, use $b-1$ top ones and column $1,2,\ldots,l+1$. Each filling is unique and has a <lattice word>, including the case $b=1$ with an empty top segment in the second filling. This proves
$$
\boxed{[a,1^{n-r-a}][b,1^{r-b}]
=[a+b,1^{n-a-b}]+[a+b-1,1^{n-a-b+1}]
+\text{non-hook constituents}.}
$$
Both displayed multiplicities are one.
To deduce the full-cycle value from this product result, rather than quote the <Murnaghan–Nakayama rule> again, form the virtual <character>
$$
F_n=\sum_{j=0}^{n-1}(-1)^j\chi^{(n-j,1^j)}.
$$
For every proper <Young subgroup> $S_m\times S_{n-m}$ and every pair of its <irreducible characters>, the hook-product calculation gives
$$
\left\langle F_n,\operatorname{Ind}_{S_m\times S_{n-m}}^{S_n}
(\chi^\alpha\boxtimes\chi^\beta)\right\rangle=0.
$$
If a factor is not a hook there are no hook constituents; if both are hooks, their two consecutive leg lengths cancel in this alternating inner product. By <Frobenius reciprocity> and the completeness of the product <irreducible characters>, $F_n$ restricts to zero on every such subgroup. Any <permutation> with more than one cycle preserves a nonempty proper subset and is conjugate into one of these subgroups. Thus $F_n$ is supported only on the class of $n$-cycles.
Its inner product with the trivial <character> is one, since only the one-row hook contributes. That class has size $(n-1)!$, so $F_n(\rho)/n=1$, giving $F_n(\rho)=n$. For any <irreducible character>, the support calculation then yields
$$
\langle F_n,\chi^\nu\rangle=\overline{\chi^\nu(\rho)}.
$$
By <character orthogonality>, the left side is $(-1)^{n-x}$ for $\nu=(x,1^{n-x})$ and zero otherwise. These are real integers, proving
$$
\boxed{\chi^\nu(\rho)=
\begin{cases}(-1)^{n-x},&\nu=(x,1^{n-x}),\\0,&\nu\text{ is not a hook}.\end{cases}}
$$
For $n=1$ the same conclusion holds directly. This derives the requested special case from the <Littlewood–Richardson rule> and <Frobenius reciprocity>.
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