I was thinking about this concept the other day, but I couldn't reach a solid conclusion.
Visualise your right hand in front of you, and put your thumb, second and third finger in a configuration such that they are all perpendicular to one another(like the right hand rule). Draw 3d axes, and allow the hand to rotate such that under any rotation, you must have the fingers lying on the axes(imagine you can keep on rotating your wrist infinitely).
Clearly , there are 24 different positions possible. My question is , firstly, do the rotations form a group, which I assume they do , and if so, what is this group isomorphic to? My guess would be a permutation group of some kind , but I would appreciate some feedback on this.
Let $\mathcal{B}=(e_1,...,e_n)$ be the standard basis of $\mathbb{R}^n$. Let $A = \mathbb{R}e_1 \cup ... \cup \mathbb{R}e_n$ be the axes. Then we need all matrices $R \in SO(n)$ with $R\mathcal{B} \subseteq A$, i.e. if $R = (r_1,...,r_n)$ with $r_i \in \mathbb{R}^n$ it has to be $r_i = \lambda_i e_i, \ i = 1,...,n$ for some $\lambda_i \in \mathbb{R}$. Since $R$ is orthogonal and has determinant $1$, the matrix $R$ is of the form $R = (\varepsilon_1 e_{\sigma(1)},...,\varepsilon_n e_{\sigma(n)})$ with $\varepsilon_i = \pm 1, \sigma \in S_n, \det R = 1$. So, $R = \text{diag}(\varepsilon_1,...,\varepsilon_n)\cdot (e_{\sigma(1)},...,e_{\sigma(n)})$ ($\text{diag}$ diagonal matrix), and the group of all these matrices is $G = \{R \in G' \ | \ \det R = 1\}$ with $G' = \{\text{diag}(\varepsilon_1,...,\varepsilon_n)\cdot (e_{\sigma(1)},...,e_{\sigma(n)}) \ | \varepsilon_i = \pm 1, \sigma \in S_n\} \subseteq O(n)$.
Now, $G' \cong (\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$, and the index $[G':G] = 2$. $G'$ has $2^n \cdot n!$ elements and therefore $G$ has $2^{n-1}n!$ elements. For $n = 3$ $G$ has $2^2\cdot 3! = 4 \cdot 6 = 24$ elements.