Background: As we all know, compared with general smooth manifolds, the biggest characteristic of Lie group is that it has a left translation mapping to itself, which is a diffeomorphism. And this makes it possible for us to use this left translation to obtain a smooth vector field on the Lie group as long as we take any tangent vector in the tangent space at the unit element. This is also the fundamental reason why every Lie group admits a global frame and therefore every Lie group is parallelizable.
My question: Consider orthogonal matrix group ${\rm O}(n)$. Suppose we already know that $T_{I_n}{\rm O}(n)$ has the following expression: $$T_{I_n}{\rm O}(n)=\{B\in {\rm M}(n,\mathbb{R}):B=-B^T\}.$$ How do we use left translation $L_A$ to compute the expression of $T_A{\rm O}(n)$?
Note: I have known some other methods to compute $T_A {\rm O}(n)$ and get its expression is $\{B\in {\rm M}(n,\mathbb{R}):BA^T=-AB^T\}$. However, those methods have certain limitations. They all take advantage of some particularity of matrix groups. But using the left translation to compute, I think we can extend this method to the general Lie group to a certain extent.
My attempt: I tried to compute the expression of $(dL_A)_{I_n}:T_{I_n}{\rm O}(n)\rightarrow T_A{\rm O}(n)$ and have got $$(dL_A)_{I_n}(B)=AB.$$ Indeed, consider smooth curve $\gamma(t)=I_n+tB$, we have $$(dL_A)_{I_n}(B)=\left.\frac{d}{dt}\right|_{t=0}L_A(\gamma(t))=\left.\frac{d}{dt}\right|_{t=0}A(I_n+tB)=AB.$$
This is a question out of curiosity. Any help would be great appreciated!
Your computation is correct ($L_A$ is the restriction of a linear map on the vector space $M_n(\mathbb R)$ to the submanifold $O(n)$, hence its derivative is again left multiplicaion by $A$). Hence the correct description of the tangent space is $T_AO(n)=\{AB:B^t=-B\}$ or if you want to get a direct characterization $\{C\in M_m(\mathbb R):C^t=-A^tCA^t\}$. To get the latter characterization you simply compute that if $B^t=-B$ and $A^t=A^{-1}$, then $(AB)^t=B^tA^t=-BA^t=-A^t(AB)A^t$ and vice versa.