Suppose I have a convex set $C\subset\mathbb{R}^n$ such that $0\in C$ and every Cauchy sequence in $C$ converges in $C$, but $C$ need not be bounded. (Actually I want unbounded $C$). Consider the set 
\begin{equation}\mathfrak{L}=\lbrace T:\mathbb{R}^n\rightarrow \mathbb{R}^n, \text{ $T$ is linear}, T(C)\subseteq C\rbrace\end{equation}
Is there any relation between extremal points (as well as exposed points) of $C$ and extremal points (exposed points) of $\mathfrak{L}$? 


I do not work in convex geometry, and so do not know whether the statement is making sense. I have asked this question [here](https://math.stackexchange.com/q/192272/28724), but did not get any response (and so I decided to ask it here). Please give suggestions and feel free to correct (and edit as well), if I am wrong and/or there is some ambiguity. Advanced thanks for any help.



Duality between extremal points and extremal maps
I'm not sure what counts as a "relation" here.
 It is tempting to think that an extreme point of $\mathfrak L$ would map at least some extreme points of $C$ to extreme points of $C$, but this is not true.
 For example, in ${\mathbb R}^2$ let $C$ be the convex hull of $(-1,-2)$, $(-1,2)$, $(2,-1)$ and $(2,1)$.
Then one of the extreme points of $\mathfrak L$ is $\pmatrix{-1/2 & 0\cr 0 &3/4\cr}$, which maps the extreme points of $C$ to $(1/2, -3/2)$, $(1/2, 3/2)$, $(-1,-3/4)$ and $(-1,3/4)$. 


If you want an example with $C$ unbounded, take the cartesian product of this with $[0,\infty)$. 


