I'm reading Kobayashi's book Transformation Groups in Differential Geometry and I don't understand a thing at page 15.
I don't understand why $U$ consists of transformation $a$ of $M$ that leave each $v\in V$ invariant and why $(*)$ is true? I've tried the definition of an morphism of $G$-strucure and in the case of $G=\{1\}$ we don't get any information.
I also know this proposition from the same book but I don't know if one can use it to answer my question: 
Where $K_x=(u^{-1})^*\text{K}$ is the tensor filed on $M$ for any $x\in M$ and any $u\in P_x.$ i.e. $u:\mathbb{R}^n \to T_xM$ a frame in $x.$
This is just the matter of definitions: A 1-structure on $M$ (which is an $n$-dimensional manifold), or a parallelism on $M$, is a collection $X_1,...,X_n$ of vector fields on $M$ which trivialize the tangent bundle $TM$, i.e. such that for every $p\in M$, $\{X_1(p),...,X_n(p)\}$ is a basis of $T_pM$. An automorphism of this structure is a diffeomorphism $f: M\to M$ such that $$ df_p(X_i(p))=X_i(f(p)), i=1,...,n, \forall p\in M. $$ (This is just the definition.) In other words, $$ f_*(X_i)=X_i, i=1,...,n. $$ Since $f_*: {\mathfrak X}(M)\to {\mathfrak X}(M)$ is a linear map, for every $$ Y= \sum_{i=1}^n a_i X_i, a_i\in {\mathbb R}, $$ we have $$ f_*(Y)=Y. $$