Suppose $G$ is a finite group and that $\rho: G\rightarrow O(d)$ is a faithful orthogonal representation, with action on $\mathbb{R}^d$ denoted $\cdot$. Let's say that $\rho$ is "strongly" angle preserving if for each $g\in G$ one has


\begin{equation}\langle g\cdot v, v\rangle = \langle g\cdot w, w\rangle \end{equation} for all $v, w$ on the (representation) sphere $S^{d-1}$. (There must be a name for such an action, but let's stick with this name and let someone edit it as desired). 


Examples of such representations would be all the finite subgroups of the classical algebras $\mathbb{R},\mathbb{C},$ and $\mathbb{H}$ under their standard embeddings in $O(1)$, $U(1)$, and $Sp(1)$. 


**Question:** Can anyone point me towards literature on any other groups admitting such a representation? 


These representations give free linear actions on spheres (since $g\cdot v=v$ forces $\langle g\cdot w, w\rangle =1$ for all $w\in S^{d-1}$), and these are classified, but it seems something stronger is required of the representation. 



Free linear group actions on spheres with "strong" angle preservation
I believe, the examples you gave are essentially the only ones. Indeed, let $\rho\colon G\to O(d)$ be a faithful irreducible “strongly angle-preserving” representation.


**Claim.** The image of $\mathbb{R}G$ under (the natural linear extension of) $\rho$ is a division algebra.


*Proof.* By linearity of the scalar product, for each $x\in\mathbb RG$ we have
$$
\langle \rho(x)v,v\rangle = \langle \rho(x)w,w\rangle,\quad v,w\in S^{d-1}.
$$


Let $A=\sum\_{g\in G} \alpha\_g \rho\_g\in M\_d(\mathbb R)$ be an element with nontrivial kernel and let $v\in\ker A = \ker A^\*A$. We then have
$$
0 =\langle A^\*Av,v\rangle=\langle A^\*Aw,w\rangle,
$$
whence $Aw=0$ for all $w\in S^{d-1}$, so $A=0$. This proves the claim.


Therefore, by Frobenius theorem on division algebras, the image $D$ of $\mathbb R G$ under $\rho$ is isomorphic to $\mathbb R$, $\mathbb C$ or $\mathbb H$, and we therefore can identify $G$ with its image in this division algebra.


The representation induces a natural scalar product on $D$ by
$$
\langle x,y\rangle:= \frac{1}{d} \mathrm{Tr}(y^\*x) = \frac{1}{d} \sum\_{i=1}^d\langle y^\*xv\_i,v\_i\rangle = \langle xv\_1,yv\_1\rangle = \langle xw,yw\rangle,\,w\in S^{d-1}
$$
and as $\rho$ is (in view of irreducibility) generated by a vector $w\in S^{d-1}$, it is naturally isomorphic to the multiplication representation of $D$ on itself (with the scalar product from above and the isomorphism sending $1\in D$ to $w$).


