This has been open for nearly nine years and is the top-rated unanswered question on Group Theory. So it deserves an answer, even though the questioner appears to no longer come round these parts.
First, something easy. Let $V$ be a $kG$-module, which we will assume is faithful.
Lemma: if $G$ possesses at most $m$ subgroups of prime order, and $|k|\geq m$ then $G$ has a regular orbit on $V$.
The proof is clear: each element of prime order centralizes a proper subspace of $V$. Since $V$ is not the union of $m$ proper subspaces, there is a point $x$ not centralized by any element of order $p$. Thus $x$ lies in a regular orbit. (You can obviously do much better than $m$ here, but this is the easiest to state.)
This deals with $k$ of characteristic $0$, of course. The best results are when $\gcd(|G|,|k|)=1$. Here we use a result of Halasi--Podoski from Every coprime linear group admits a base of size two. This states that any faithful module has base size $2$, i.e., there exist two points, $x,y\in V$, such that $C_G(x)\cap C_G(y)=1$. Thus if $W$ is a faithful $kG$-module and $W\oplus W$ is a submodule of $V$ then $G$ has a regular orbit on $W\oplus W\leq V$.
Thus if $V$ is a faithful $kG$-module then $V\otimes V$ contains two non-isomorphic faithful modules, namely $\Lambda^2(V)$ and $S^2(V)$. The tensor product of the two contains two copies of $\Lambda^2(V)\otimes S^2(V)$, so there is always a regular orbit in $V^{\otimes 4}$.
If $V$ is a permutation module then $V=1\oplus W$, and $W\oplus W\leq V\otimes V$. Hence there is a regular orbit in $V^{\otimes 2}$. This of course generalizes to $V$ an arbitrary sum of modules, but we have to be a little careful to make sure that $V\otimes V$ is faithful.
I would guess that even when $V$ is simple that $G$ has a regular orbit on $V\otimes V$, but this has not been proved, as far as I know.
Having dealt with the coprime case, we look at the general case. Here the base size can be greater than $2$. Some recent papers of Melissa Lee give bounds on base sizes for quasisimple groups, but as far as I know this looks quite hard. Note that a result of Auslander--Carlson states that if $V$ is absolutely indecomposable and of dimension divisible by $p=\mathrm{char}(k)$, then $V\otimes V^*\otimes V$ contains $V\oplus V$. This means that in that situation we do obtain two summands isomorphic to $V$ in $V^{\otimes 3}$, whenever $V$ is self-dual. Thus we may exploit base sizes again. In general, of course, knowing composition factors of $V^{\otimes n}$ doesn't help you too much, because you need composition factors of the socle to guarantee that you achieve a regular orbit.
My guess would be that if $V=W_1\oplus W_2$ and each $W_i$ has base size $2$, then with few exceptions $V$ possesses a regular orbit.
Take a point $p = (a,b) \in \mathbb{R}^2$. The claim is that taking any point $q =(x,y)\in\mathbb{R}^2$, we find a linear transformation (i.e. matrix) $A$ such that $A p^T = q^T$ (1).
Denote $A = \begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$. The equation (1) corresponds to the system $\alpha a + \beta b = x \,\& \, \gamma a + \delta b = y$, which has in general infinitely many solutions; adding the condition of regularity of $A$, i.e. the zero determinant, will cancel some of them.
Best Answer
$1$.) I count 16 as well:
$D_{12}$ itself and the identity (trivial) subgroup.
$\langle r\rangle,\langle r^2\rangle, \langle r^3\rangle$
Two isomorphs of $S_3$: $\langle r^2,s\rangle$ and $\langle r^2,rs\rangle$.
Three isomorphs of $V$: $\langle r^3,s\rangle$, $\langle r^3,rs\rangle$ and $\langle r^3,r^2s\rangle$.
Six subgroups generated by a single reflection $\langle r^ks\rangle$ for $k = 0,1,2,3,4,5$.
$2$.) Yes, you want to find the normalizers in $D_{12}$. This is the entire group (as these are all normal) for the first seven subgroups listed, which greatly simplifies things. The isomorphs of $V$ are non-normal but of prime index, so they must be their own normalizers. It is clear that the normalizers of the reflection-generated subgroups of order 2 must be an isomorph of $V$ (the isomorphs of $V$ are abelian so they normalize any subgroup), as $S_3$ has no normal subgroups of order 2 (we are using the fact that the normalizer of a subgroup $H$ in a group $G$ is the largest subgroup of $G$ containing $H,$ in which $H$ is normal).
$3$.) Finding the orbits by brute force is not so bad: the seven normal subgroups will all have a single element of just themselves in their orbits, because they are stabilized by all of $D_{12}$. From Sylow theory (or by inspection), we see the isomorphs of $V$ are all conjugate, so there is another orbit (remember conjugation preserves order of subgroups), and we are left with finding the orbits of the reflection-generated subgroups.
By the orbit-stabilizer theorem (or by examining their generators' (ordinary) conjugacy classes) we have that these must occur in two orbits of three: $\{\langle s\rangle,\langle r^2s\rangle,\langle r^4s\rangle\}$ and $\{\langle rs\rangle,\langle r^3s\rangle,\langle r^5s\rangle\}$
Thus: $G/X = \{\{\{1\}\},\{\langle r\rangle\},\{\langle r^2\rangle\}, \{\langle r^3\rangle\},\{D_{12}\},\{\langle r^2,s\rangle\},\{\langle r^2,rs\rangle\},\{\langle r^3,s\rangle, \langle r^3,rs\rangle,\langle r^3,r^2s\rangle\},\{\langle s\rangle,\langle r^2s\rangle,\langle r^4s\rangle\},\{\langle rs\rangle,\langle r^3s\rangle,\langle r^5s\rangle\}\}.$
(This is a set with 10 elements).