TITLE: Tangent Space of $\operatorname{Aut}(T_e G)$
QUESTION [5 upvotes]: Let $G$ be a Lie Group, how to prove that the tangent Space of $\operatorname{Aut}(T_e G)$ at identity is $\operatorname{End}(T_e G)$? Thanks.

REPLY [2 votes]: If $V$ is a finite-dimensional vector space, considered as a manifold, then the tangent space is $V$ at any point. And $\mathrm{Aut}$ is an open subspace of the vector space $\mathrm{End}$. This doesn't have anything to do with Lie groups.