Let $G$ be a lie group, $g\in G$. We have
$$C_g:G \longrightarrow G$$ $$C_g(h) = ghg^{-1}$$
Since $C_g$ is diffeomorphic, we know that $$(C_g)_{*e}: T_eG \longrightarrow T_eG$$ is an isomorphism, hence $(C_g)_{*e} \in L(T_eG)$.
Let $F: G \longrightarrow L(T_eG)$, $F(h) = (C_h)_{*e}$
Since $L(T_eG)$ has a vector space structure, it obviously is a smooth manifold. But the problem is how do I show that $F$ is smooth?
First observe that smoothness of $F$ is equivalent to smoothness of the corresponding map $G\times T_eG\to T_eG$. Now start from the observation that $\phi(g,h):=ghg^{-1}$ is smooth as a map $G\times G\to G$. Hence the tangent map $T\phi:T(G\times G)\to TG$ is smooth. Next, $T(G\times G)$ can be identified with $TG\times TG$ by the tangent maps of the projections onto the two factors. Since the zero vector field $g\mapsto 0_g$ and the inclusion $T_eG\to TG$ are evidently smooth maps, you get a smooth map $G\times T_eG\to TG\times TG$ defined by $(g,X)\mapsto (0_g,X)\in T_{g,e}(G\times G)$. Hence $(g,X)\mapsto T_{(g,e)}\phi(0_g,X)$ is a smooth map $G\times T_eG\to TG$. But since $\phi(g,e)=e$ for all $g$, this has values in $T_eG$ and hence is smooth as a map to this space. But from the construction it is clear that $T_{(g,e)}\phi(0_g,X)$ just the derivative of $C_g$ applied to $X$.