Invertible (isometric) sections of certain hom bundles over sphere Assume  that we  have  a  vector  bundle $E$ over  $S^n$.

Is  there  a  continuous  family of invertible linear  maps  $T_x:E_x \to  E_{-x}$?

Here  continuity has  the  obvious  meaning as soon as  we  have  trivialization for  the  bundle  around $x$ and  $-x$
There is an obvious affirmative answer for the non orientable line bundle over $S^1$,  for the tangent bundle of $S^n$ and  for  the  $2$  dimensional real vector  bundle $E \to S^2 \simeq \mathbb{C}P^1$ where  $E$ is the  tautological  complex line  bundle  over  $\mathbb{C}P^1$.
If the  answer is  yes, what  about  if  we  consider  $E$  as  a  Riemannian  bundle  and  we  require that every  $T_x$  be an  isometric  linear  map?
 A: There is no such family of complex linear maps for the tautological bundle on $\mathbb CP^1$. The antipodal map is orientation reversing, hence the first Chern class changes its sign. The same argument should work for the positive complex spinor bundle on any even-dimensional sphere.
The same argument also works for a real vector bundle on $S^{4k}$ with non-vanishing $k$-th Pontryagin class. Probably again the positive real spinor bundle is an example.
On the other hand, the answer to the second question is "yes" because $O(r)$ or $U(r)$  is a deformation retract of $GL(r,\mathbb R)$ or $GL(r,\mathbb C)$, respectively.
A: You are asking if a real vector bundle of rank $k$ over $S^n$ is isomorphic to its antipodal image.
If $n$ is odd, the answer is always yes because the antipodal map is homotopic to the identity of $S^n$.
If $k=2$, the answer is easily seen to be yes, because the oriented $2$-bundles are classified by an integer, the Euler class.
If $n$ is even, such a real vector bundle amounts to a free homotopy class of continuous maps $S^n\to BO(k)$. i.e. to a pair of elements $\alpha, \bar\alpha$ of the homotopy group
$\pi_n(BSO(k))$, where $BSO(k)$ is the Grassmannian space of the oriented $k$-planes in $R^\infty$, and where $\bar\alpha$ is the image of $\alpha$ by an orientation-reversing isometry. One has $\pi_n(BSO(k))\cong\pi_{n-1}(SO(k))$. Your question amounts to ask that $\alpha=\alpha^{-1}$ or $\alpha=\bar\alpha^{-1}$. I believe that the second always holds, but I'm not sure.
