Combinatorics for a 3-d rotating automaton 
*

*Let's suppose that we have some kind of special 3-dimensional
rotating automaton.


The automaton is capable to generate rotation about selected  $X$ or $Y$ or $Z$ axis (in a current frame) in steps by only constant  +$\dfrac{\pi}{6}$ angle (i.e. rotation can be generated only in one direction - reverse rotation is prohibited) so  transition from matrix $R_{i-1}$ to  $R_{i}$ (right-lower indices denote here states before and after a single step) is achieved with the use of formula:  
$R_{i}=Rot_{x,y,z}( \dfrac{\pi}{6})R_{i-1}$  
Initial state is coded as the identity matrix $R_0=I$, all other states are described as rotation matrices in reference to the frame representing by this  $I$ matrix.
Questions:


*

*how many $n$ distinct states (coded in generated matrices) can be
achieved for  not limited number of steps.  This full set of achievable states coded $\{^{1}R ,^{2}R ...{^{n}R} \}$ might be named to be a full space of rotating automaton (here left-upper indices should be somehow reasonably organized, but hard to say how - it's open issue) - all states and transitions between states can be, perhaps, visualized with the use of a graph

*how many distinct states can be generated by exactly $6$ steps in
the automaton (...maybe there is a general formula for $n$ steps ?)

*by how many ways can be achieved multi-step rotation from $I$ to $I$
with the condition that on this trajectory of states the same one step  transition ${^{j}R}{\rightarrow}{^{k}R}$ (if possible) is allowed only one time.
(for example if it were only rotation about a single axis allowed - the number would be obviously $3$ i.e. three 12-step transitions, but in general case rotations about different axes can be mixed)
 A: Note: this is a barely-better-than-nothing answer!


*

*How many different positions can be generated given an unlimited number of steps? The answer is: an infinite number of positions. I believe I have a proof, but it appears long and involved and full of potential pitfalls. But the essence lies in the impossibility of having a closed curve on the sphere if you alternate between a $\pi/6$ rotation along one axis, a rotation along a second axis, the "reverse" of the first (which you obtain with $11$ steps in the other direction) and the "reverse" of the second: think "up", "left", "down", "right". I believe anyone reasonably well-versed in spherical trigonometry should be able to show this.
In fact, I believe one can prove that, even if one can use only $2$ of the admissible rotations, and actually for any "step" that is an angle other than an integer multiple of $\pi/2$, for any final $R$ and any arbitrarily small $\epsilon$ there is a sequence of rotations that brings the robot to some $R'$ such that the maximum difference between any element of $R$ and $R'$ is less than $\epsilon$.

*How many distinct positions can you generate with $6$ steps? Clearly no more than $3^6$. I believe that it's actually $3^6-3$, since based on the same arguments of point $1$ above the only possibility of ending up in the same position with a sequence of $6$ steps is to take two different sequences each resulting in a $\pi/2$ rotation (so, $3$ identical steps, followed by $3$ identical steps). Out of the $9$ possible cases, $3$ (with "identical" subsequences) lead to distinct positions, the remaining $6$ lead form $3$ pairs that lead to $3$ distinct positions.

*"By how many ways can be achieved multi-step rotation...?" I am not sure I understand the question correctly, but I believe that the answer is infinite, based on $1$. The idea is this: assume a "generalized move" $R$ is a sequence of $s$ identical rotation steps around the same axis, with $1\leq s \leq 11$. Denote by $-R$ the "reverse" of $R$, a sequence of $12-s$ identical steps around the same axis: obviously applying first $R$ and then $-R$ brings you back to the starting position without ever being in the same state twice in-between. Now assume you can reach from I a position R, and consider a sequence of generalized moves $R_1,...,R_n$ that is "minimal", i.e. there is no shorter sequence of generalized moves from I to R. Then the sequence $R_1,...,R_n,-R_n,...,-R_1$, brings you back to $I$, and never performs the same sequence of rotations between two identical positions twice.
Note: I've abstracted and generalized questions 1 and 3 in the related set of questions that you can find here!
