A cone is a $R\_+$-module. That is, a cone is an abelian monoid that is closed under nonnegative real scalar multiplication. An automorphism of a cone is a bijective $R\_+$-linear map. That is a map $f:C\to C$ such that $f(\alpha x +\beta y)=\alpha f(x) + \beta f(y)$ for all $\alpha,\beta \ge 0 $ and $x,y\in C$.


Suppose I have an inclusion of cones $C\subset D$. If the automorphism group $Aut(C)$ and $Aut(D)$ are isomorphic, does that imply that $C=D$? You may assume that these cones are finite-dimensional, in the sense that there is an $R\_+$-linear map from each of these cones to some euclidean space $R^d$.



Does the automorphism group of a cone determine the cone?
No. Consider solid angles in $\mathbb R^2$. They all (except the half-plane) are isomorphic, yet one may be strictly included in another.


Furthermore, a generic cone in $\mathbb R^d$, $d\ge 3$, has a trivial automorphism group, so you may have $Aut(C)=Aut(D)$ in the strongest possible sense but $C\ne D$.


