A (possible) equivalent relation on the space of vector bundles Edit: According to the essential comment of Alex Degtyarev, we revise the question as follows;

Assume that $\alpha$ and $\beta$ are two  oriention preserving automorphism of Lie groups $O(n)$ and $O(m)$, respectively. Is there an oriention preserving automorphism of $O(n+m)$ which restriction to $O(n)\oplus O(m)$ equal to $\alpha \oplus \beta$? 

One can repeat the same question for complex version,i.e:$U(n)$.
The reason for consideration of "oriention preserving" is that: It is unlike that the  answer would be positive for $\alpha(z)=z$  and $\beta(w)=\bar{w}$, as automorphisms of $U(1)$.
The reason for  consideration of $O(n)$, instead of $GL(n,\mathbb{R})$ is that: it is unlike that the answer would be positive for $\alpha(x)=x^{3},\;\beta(x)=x$, as automorphism of $GL(1,\mathbb{R})$.
Our main motivation:
We try to define  a (possible) equivalent relation on Riemannian vector bundles as follows:
Assume that $E$  and  $F$  are two vector  bundles over $X$.
We say that $E$ is  equivalent to $F$ if there are   cocycles $g_{\alpha\beta}$ and $h_{\alpha\beta}$, respectively  for  $E$  and  $F$   and  an (oriention preserving) automorphism $\lambda$ of $O(n)$ such that $g_{\alpha\beta}=\lambda\circ h_{\alpha\beta}$.

Is this really  an equivalent relation?(Transitive property?) Is there a  sheaf theoretical language for this question(for this relation)?

Of course, every two ordinary isomorphic bundles, are equivalent in this definition.
If the  answer is yes, We wish to define the direct product on this  structure (on the  space of  all equivalent class of Riemannian vector  bundles) to obtain a semigroup. So the  above  question is needed for "well define-ness". Then we  consider the Grothendieck  group of this  semigroup. So we  would have a functor $\tilde{\tilde{K}}(X)$. 

Now perhaps a natural question is that what would be  an  appropriate  version of Atiyah   Janich theorem, here? 

As Alex Degtyarev commented on previous version of this post, we  ask that how can we define an  appropriate higher order of  $\tilde{\tilde{K}}$?
 A: EDIT: This is now a full answer, and the answer is "no".
Let $G$ be a structure group as in the question.
Let $\lambda$ and $\mu$ be automorphisms, then so is $\lambda\circ\mu$.
We write $E\sim_\lambda F$ if  $E$, $F$ have cocycles $g_{\alpha\beta}$, $h_{\alpha\beta}$ respectively such that $g_{\alpha\beta}=\lambda\circ h_{\alpha\beta}$. The following things are easy to check.


*

*$\sim_\lambda$ is compatible with the cocycle condition.  

*If $g'_{\alpha\beta}=c_\alpha g_{\alpha\beta}c_\beta^{-1}$ is an equivalent cocycle for $E$, then the $0$-cochain $(\lambda\circ c_\alpha)$ can be used to find an equivalent cocycle $h'_{\alpha\beta}$ for $F$ to which $g'_{\alpha\beta}$ is again $\lambda$-related.

*$E\sim_\lambda F$ if and only if $F\sim_{\lambda^{-1}}E$.

*$E\sim_\lambda F$ and $F\sim_\mu D$ imply $E\sim_{\lambda\circ\mu}D$ (this needs step 2).

*If $\lambda$ is an inner automorphism induced by an element $g_0\in G$
and $g_{\alpha\beta}$ is $\lambda$-related to $h_{\alpha\beta}$, then $g_{\alpha\beta}$ and $h_{\alpha\beta}$ are in fact equivalent cocycles related by the constant $0$-cochain $g_0$.
Define $E\sim F$ if and only if there exist cocycles describing $E$ and $F$ which are $\lambda$-related for some $\lambda$. Then $E\sim F$ is the same as $E\cong F$ unless $G$ has outer automorphisms.
Outer automorphisms of simply connected, connected Lie groups can be read off from the Dynkin diagram. Neither $O(n)$ nor $U(n)$ is of this type, but you can still make some deductions.
For example, the central $S^1$ in $U(n)$ contains $n$ elements of $SU(n)$, which determines the action of outer automorphisms on the center for $n\ge 3$.
Hence the group $U(n)$ has one (equivalence class of nontrivial) outer automorphisms given by complex conjugation (even for $n=1$, $2$). If $E\sim_\lambda F$, then
$$c_1(E)=\pm c_1(F)$$
with "$-$" if and only if $\lambda$ is outer.
Hence there is no way to extend identity on $U(m)$ and conjugation on $U(n)$ to an automorphism of $U(m+n)$ because in general
$$c_1(E\oplus\bar F)\ne\pm(c_1(E)+c_1(F))\;.$$
So you cannot define your version of complex $K$-theory that way.
All outer automorphisms of $SO(n)$ come from inner automorphisms of $O(n)$ (the triality automorphisms of $Spin(8)$ do not act on $SO(8)$). The group $O(n)$ is generated by reflections, and each reflection is specified up to sign by its conjugation action on $SO(n)$ for $n\ge 3$. So the only possible outer automorphism of $O(n)$ is given by $\lambda(g)\mapsto g\cdot\det g$ in the case that $n$ is even (for $n=2$, this is actually an inner automorphism as well). If $E\sim_\lambda F$ then $F\cong E\otimes o(E)$, where $o(E)=\Lambda^nE$ denotes the orientation bundle. In particular, if $E$ is orientable, then $E\cong F$.
Using the splitting principle, one computes the Stiefel-Whitney classes
$$w_k(E\otimes o(E))=\sum_{i=0}^k\binom{n-i}{k-i}w_i(E)\,w_1(E)^{k-i}\;.$$
Already for $n=4$ and $k=2$, one has $w_2(E\otimes o(E))=w_2(E)+w_1(E)^2$,
so the bundles $E$ and $E\otimes o(E)$ are not stably isomorphic.
To come back to the original question: for $m$ odd, the bundles $E\oplus\underline{\mathbb R}^m$ and $(E\otimes o(E))\oplus\underline{\mathbb R}^m$ are not diffeomorphic, and because $m+n$ is odd, there is no outer automorphism to relate them. So again, you cannot define a new $K$-group.
