Let us remember the definition of faithful, transitive and free:
The action of a group $G$ on a set $X$ is called:
Transitive: if $X$ is non-empty and if for any $x,y\in X$ there exists $g\in G$ such that $g\cdot x = y$.
Faithful (or effective): if for any two distinct $g,h\in G$ there exists $x\in X$ such that $g\cdot x \neq h\cdot x$; or equivalently, if for any $g\neq e\in G$ there exists $x\in X$ such that $g\cdot x \neq x$. Intuitively, in a faithful group action, different elements of $G$ induce different permutations of $X$.
Free: if, given $g,h\in G$, the existence of $x\in X$ with $g\cdot x = h\cdot x$ implies $g = h$. Equivalently: if $g$ is a group element and there exists an $x\in X$ with $g\cdot x = x$ (that is, if $g$ has at least one fixed point), then $g$ is the identity.
Now we can determine your three group actions:
The action of $S_n$ on a set of $n$-elements is as you showed transitive. It is also faithful as you explained, but it is not free, because there are elements on $S_n$ that do have fixed points that are not the identity, for example $1\to 1, 2\to 3, 3\to 4, n-1\to n, n\to 2$. Since the action is transitive the group of orbits is the group of one element.
The action of $O(n)$ on $S^{n-1}$, it is transitive, since there is always a rotation that can send any point of $S^{n-1}$ onto the north pole. For $n>2$, It is not free since for any point on $S^{n-1}$ there are many rotations different from the identity, whose axis of rotation is the line that passes through that point and the origin, and all of these fix that point. It is not faithful neither since any element of $O(n)$ that is not the identity moves $S^{n-1}$, therefore there exists a point on $S^{n-1}$ that is not fixed by the action. Since the action is transitive the group of orbits is the group of one element. The case $n=2$ the action it is transitive, free but not faithful.
The action of $O(n)$ on $\mathbb{R}^n$ is not free since every element of $O(n)$ fixes the origin but is not necessarily the identity. It is not transitive either, since for two elements of $\mathbb{R}^n$ whose length are not equal, you cannot find an element of $O(n)$ that sends one on the other (the elements of $O(n)$ preserve norm). It is not faithful neither since when you restrict the action to $S^{n-1}$ it is not faithful either. Since $\mathbb{R}^n=[0,\infty)\times S^{n-1}$ the set of orbits can be identified with $[0,\infty)$ since when $O(n)$ is restricted to the second factor it is transitive.
Hint: The number of distinct $k$-cycles is $P^n_k\cdot \dfrac 1k=\dfrac{n!}{(n-k)!}\cdot \dfrac 1k$.
To do your example, we would get, for permutations of type $(2,3,2)$ in $S_{15}$:
$P^{15}_2\cdot \dfrac 12\cdot P^{13}_3\cdot \dfrac 13\cdot P^{10}_2\cdot \dfrac 12=105\cdot572\cdot45=2702700$.
Now I need to divide by $2$, since I have double counted the two $2$-cycles.
So $\dfrac12\cdot2702700=1351350$.
See here, or here, for a good explanation.
Best Answer
I cannot find the proof you are looking for in the book by D. E. Taylor that ego suggested in the comments either. But you can use the techniques of the section "Reflections and the Strong Exchange Condition" on pages 94-96 for a proof (my answer is independent of the book, for connections see the comment at the end).
In your case you have the set $R = \{(1\;-1)\} \stackrel{.}{\cup} \{(i\;i+1)(-i\;-i-1) \mid i \in [n-1]\}$ of involutions generating the Coxeter group $W = \langle R\rangle$, which is a subgroup of the symmetric group on $[n] \cup -[n]$ (using the notation $[n] := \{1, \dots, n\}$).
For the set $T = \{wrw^{-1} \mid w \in W, r \in R\}$ of all conjugates of the generating involutions in $R$ (defined in equation (9.20) of the book) you should be able to check $T = T_1 \stackrel{.}{\cup} T_2 \stackrel{.}{\cup} T_3$ for $$T_1 := \{(i\;-i)\mid i\in[n]\}$$ $$T_2 := \{(i\;j)(-i\;-j) \mid i, j \in [n], i<j\}$$ $$T_3 := \{(i\;-j)(-i\;j) \mid i, j \in [n], i<j\}$$ $T$ consists of two different conjugacy classes: $T_1$ are the conjugates of $(1\;-1)$), and $T_2\stackrel{.}{\cup}T_3$ the conjugates of all the other elements of $R$.
$W$ acts on $T$ by conjugation, which induces an action on the power set $\mathcal{P}(T)$ of $T$. With the symmetric difference as addition, the power set $\mathcal{P}(T)$ becomes an elementary abelian $2$-group, on which $W$ acts. Define $D'(w) = D'_1(w) \stackrel{.}{\cup} D'_2(w) \stackrel{.}{\cup} D'_3(w)$ for $w \in W$ with $$D'_1(w) = \{(i\;-i) \in T_1 \mid i\in [n] \mbox{ and } w(i)<0\}$$ $$D'_2(w) = \{(i\;j)(-i\;-j) \in T_2 \mid i, j\in [n], i<j\mbox{ and } w(i)>w(j)\}$$ $$D'_3(w) = \{(i\;-j)(-i\;j) \in T_3 \mid i, j\in [n], i<j\mbox{ and } w(i)+w(j) < 0\}$$ The cardinality of $D'(w)$ is just inv$(w)$ from your question.
Claim: (a) $D'(r) = \{r\}$ for $r \in R$
(b) $D'(w_1w_2) = w_2^{-1}D'(w_1)w_2+D'(w_2)$ for $w_1, w_2 \in W$
The first statement is easily verified, we show the second one:
For this, observe that an element $(i\;-i)$ of the conjugacy class $T_1$ is contained in $D'(w)$ if and only if $\frac{w(i)}{i}<0$ for $i \in [n]\cup-[n]$.
Now $(i\;-i) \in D'(w_1w_2)$ is the same as $0 > \frac{w_1(w_2(i))}{i} = \frac{w_1(w_2(i))}{w_2(i)}\cdot \frac{w_2(i)}{i}$, which in turn is equivalent to either $ D'(w_1) \ni (w_2(i)\;-w_2(i)) = w_2(i\;-i)w_2^{-1}$ or $(i\;-i) \in D'(w_2)$, i.e., $(i\;-i) \in w_2^{-1}D'(w_1)w_2+D'(w_2)$.
An element $(i\;j)(-i\;-j)$ of the conjugacy class $T_2\cup T_3$, where $i, j\in [n]\cup -[n]$, $i\ne j$ and $i\ne -j$, is contained in $D'(w)$ if and only if $\frac{w(j)-w(i)}{j-i}<0$ (check this by considering the two cases $i\cdot j>0$ and $i\cdot j<0$ and by considering what happens if you replace $i$ and $j$ by $-i$ and $-j$).
So $(i\;j)(-i\;-j) \in D'(w_1w_2)$ is equivalent to $0 > \frac{w_1(w_2(j))-w_1(w_2(i))}{j-i} = \frac{w_1(w_2(j))-w_1(w_2(i))}{w_2(j)-w_2(i)}\cdot \frac{w_2(j)-w_2(i)}{j-i}$, which is the same as $(i\;j)(-i\;-j) \in w_2^{-1}D'(w_1)w_2+D'(w_2)$.
Having proven the claim, one can easily deduce the formula for the length by induction:
Because of (a) we may assume that $l(w) > 1$, and write $w = s\cdot v\cdot t$ with $s, t \in R$ and $l(v) = l(w)-2$. Per induction $|D'(v)|+1 = l(v)+1 = l(v\cdot t) = |D'(v\cdot t)| \stackrel{(b)}{=} |tD'(v)t + D'(t)| \stackrel{(a)}{=} |tD'(v)t + \{t\}|$, i.e., $t \not\in D'(v)$. As $v\ne w = svt$ we get $t \ne v^{-1}sv$, and hence $t\not\in \{v^{-1}sv\} + D'(v) \stackrel{(a)}{=} v^{-1}D'(s)v + D'(v) \stackrel{(b)}{=} D'(s\cdot v)$. It follows $D'(w) \stackrel{(b)}{=} tD'(s\cdot v)t \stackrel{.}{\cup} \{t\}$ and therefore the induction step.
The claim is a variant of (9.22) in Taylor's book:
For $D(w) := wD'(w)w^{-1}$ you get $D(r) = r\{r\}r = \{r\}$ as $r$ has order 2. Also $D(w_1w_2) = w_1D'(w_1)w_1^{-1}+w_1w_2D'(w_2)w_2^{-1}w_1^{-1} = D(w_1)+w_1D(w_2)w_1^{-1}$ showing that D fulfills (9.22).
This condition on $D$ is quite powerful. Taylor uses it to derive the strong exchange property (with Corollary 9.26 implying the formula for the length), and that it is equivalent to $(W, R)$ being a Coxeter system.