Number of orbits of $G$ acting on $X$ This question comes from Algebraic Combinatorics: Walks , Trees, Tableaux, and More by Richard P. Stanley. It is written as follows:
"Let $X$ be a finite set, and let $G$ be a subgroup of the symmetric group $S_{X}$. Suppose that the number of orbits of $G$ acting on $n$-colorings of $X$ is given by the polynomial
$$f(n)=\frac{1}{443520}\left(n^{11}+540n^{9}+...+10n\right).$$
How many orbits does $G$ have acting on $X$?"
I am interested in determining the number of orbits of $G$ acting on $X$. I thought the answer was $1$ since the presence of the term $10n$ seems to indicate that there is a permutation that has only one cycle so I thought it must contain all elements of $X$ since we even count $1$-cycles. To me this meant we could travel from any element of $X$ to any other element by means of this cycle. Apparently the answer is $10$ and I was very wrong. I was hoping that someone could help clear this up. Do I perhaps misunderstand the question or simply do not have my definitions correct? Thank you for your help.
 A: Edit: In the first part you'll find my own attempts that almost answer the question. Later, I decided to read the corresponding chapter in the book and give a more concise and full solution. I still believe the first part will be useful to anyone who hasn't read Stanley's book.

First of all, let's see if we understand the problem the same way:
$G$ acts on two different sets; one is $X$, the other is the set of $n$-colorings of $X$ (the latter depending on $n$ and being much bigger after a while). It is important to understand that the orbits of $X$ are not affected by any coloring... They are there to begin with; when we consider colorings, we allow $G$ to act on that different set...
Now, an important lemma: If two colorings $C_1,C_2$ of $X$ involve different colors in the same orbit of $X$ (that is, there is an orbit $X_i$ of elements of $X$ and in $C_1$, the color blue is used in $X_i$ but it is not used, again in $X_i$, in the coloring $C_2$), then... the two colorings $C_1,C_2$ cannot belong to the same orbit under $G$ (in the set of colorings!).
Now, it is easy to see that any coloring $C_i$ for which, for all orbits $X_i$ of $X$, all points in $X_i$ have the same color, will form an orbit on its own in the set of colorings. Indeed, the group $G$ will act permuting elements in each orbit but that would give the same coloring since the color is constant in each orbit. 
This implies that if we have $k$-many orbits in $X$, we will have at least $n^k$ orbits in the set of colorings. That means in our example that $k\leq 10$.
Let's be more precise: Say there are $k$-many orbits with sizes $s_1,s_2,\cdots,s_k$ (so that $s_1+\cdots+s_k=|X|$ ). Then, any orbit in the set of colorings can have at most $s_1!\cdot s_2!\cdots s_k!$ elements. That means there are at least $\displaystyle \frac{n^{|X|}}{s_1!\cdot s_2!\cdots s_k!}$-many orbits. But of course, any orbit has at least one element, so there are at most $n^{|X|}$-many orbits. These two statements together imply that $|X|=11$.
Now, we know evenmore! We know that if the orbit sizes are $s_1,\cdots,s_k$ as above, we need $$\frac{n^{11}}{443520}\geq \frac{n^{11}}{s_1!\cdot s_2!\cdots s_k!}$$ This in turn means $$s_1!\cdot s_2!\cdots s_k!\geq 443520=8!\cdot 11$$ Now, given that $s_1+\cdots+s_k=11$, the previous inequality is only true when $s_1=9$ and $s_2=2$, or when $s_1=10$ and $s_2=1$, or when $s_1=11!$.
This implies that we might have two orbits or only one.

I decided to go through the book a little bit and I found out that the following theorem is provided for the number $N_G(n)$ of inequivalent $n$-colorings: $$N_G(n)=\frac{1}{\#G}\sum_{\pi\in G}n^{c(\pi)}$$ where $c(\pi)$ is the number of cycles of $\pi$. Now, I would first urge you to include such information in future questions, as the audience might not be familiar with any result in Combinatorics :). Then, I believe this theorem agrees with your opinion and what I've been writing above:
Since, the identity is always part of the group $G$ and has $|X|$ many cycles, we may deduce both that $|X|=11$ and that $\#G=8!\cdot 11=443520$. Then, the term $10n$ implies that there are $10$ permutations with only one cycle. This implies that there exists exactly one $G$-orbit in $X$ (that is, $G$ acts transitively on $X$). It makes sense to have ten such permutations since the cyclic group $C_{11}$ has ten elements that generate it.
