I'm reading Kobayashi's book, Transformation Groups in Differential Geometry and at the page 3 is this proposition: 
the definition of $K$ is given here:
My question is why this proposition is evident? I know how to prove $(2)$ using $(1)$ Any hint will help, I'm sure.
A $G$-structure defined on $M$ is a $G$-reduction $P$ of the bundle of linear frames $L(M)$. Suppose that the $G$-structure is defined by the tensor $K$. An automorphism of $P$ is a diffeomorphism of $M$ such that $f^*P=P$.
Let $x\in M$, and $u:\mathbb{R}^n\rightarrow T_xM$ and element of $P_x$, $f$ is an automorphism of $P$ if and only if for every $x\in M, u\in P_x$, $df_x\circ u:\mathbb{R}^n\rightarrow T_{f(x)}M$ is an element of $P_{f(x)}$. Suppose that $K$ is a $p$-tensor. This is equivalent to the fact that for every $v_1,...,v_p\in T_{f(x)}M$, $K_{f(x)}(v_1,...,v_p)=K((df_x\circ u)^{-1}(v_1),...,(df_x\circ u)^{-1}(v_p))=K(u^{-1}(df_x^{-1}(v_1)),...,u^{-1}(df_x^{-1}(v_p)))=K_x(df_x^{-1}(v_1),...,df_x^{-1}(v_p))$.
This is equivalent to the fact that $f$ preserves the tensor defined on $M$ by $K$.