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According to the answer of Sebastan and previous edit of Ben McKay I revise my post as follows:

Assume that $E$ is a vector bundle over a manifold $M$ with a connection $\nabla$. Is there a (unique) connection $\nabla'$ on $E':=Hom (E,E)$ with the following property;

For every curve $\gamma$ which connects point $x$ to $y$, with $\nabla$ parallel transport $\phi$ and $\nabla'$ parallel transport $\psi$, we have $\psi(T)=\phi T\phi^{-1}$. Moreover if $\nabla$ is a Riemannian connection corresponding to a Riemannian metric on $E$, can we choose a Riemannian comnnection $\nabla '$ as above.In the latter we consider the natural Riemannian metric inducing by initial metric on $E$ defined by $tr(AB^{*})$ on the $E'$- bundle

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  • $\begingroup$ @Ben thank you for your edit. It seems that we were editing simultanously. In my new edit I fix my errors you pointed out. $\endgroup$
    – Ali Taghavi
    Commented Jul 4, 2016 at 11:34
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    $\begingroup$ Perhaps it should be pointed out that this is a special case of a more general fact: Each connection $\nabla$ on $E$ induces a canonical connection $\nabla^{r,s}$ on $E^{\otimes r}\otimes (E^*)^{\otimes s}$ that is compatible with all tensor products, contractions, sub-representations, etc., so all of the parallel transports are also compatible. Your case is just $(r,s)=(1,1)$. $\endgroup$
    – Robert Bryant
    Commented Jul 4, 2016 at 17:21
  • $\begingroup$ @RobertBryant Thank you very much for your comment. And the Riemannian property would be preserved? $\endgroup$
    – Ali Taghavi
    Commented Jul 4, 2016 at 18:38
  • $\begingroup$ Yes, certainly. $\endgroup$
    – Robert Bryant
    Commented Jul 4, 2016 at 20:34

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No, because your formula does not make sense:

$T\in Hom(E_x,X_x)$ and $\phi\in Hom(E_x,E_y)$ invertible means that $$\phi^{-1}\circ T\circ \phi$$ is not well-defined unless $x=y.$

If you define $$\psi=\phi\circ T\circ \phi^{-1},$$ then $\psi$ is actually the parallel-transport of the induced connection $\nabla^{End}$ on the endomorphism bundle which is defined by satisfying the equation $$(\nabla^{End}T)(e)=\nabla T(e)-T(\nabla e)$$ for all $T\in\Gamma(Hom(E,E))$ and sections $e\in\Gamma(E).$

For a proof of this property consider $v,w\in E_x$ with $T(v)=w,$ and denote the corresponding parallel sections (along the given curve $\gamma$ from $x$ to $y$) by $v(t)$ and $w(t).$ Then, $\nabla_{\gamma'} v(t)=0$ and $\nabla_{\gamma'}w(t)=0.$ Hence, by uniqueness of solutions of ODE's, the parallel endomorphism field $T(t)$ along $\gamma$ satisfies $$T(t)(v(t))=w(t).$$ As $v\in E_x$ is arbitrary, this is equivalent to the equation $\psi=\phi\circ T\circ \phi^{-1}.$

The construction is compatible with metrics, as the same standard arguments for tensor products, dual bundles and corresponding connections carry over to the Riemannian/hermitian situation, i.e., the induced metric on the endomorphism bundle is parallel with respect to $\nabla^{End}.$ Note also that $\nabla^{End}$ is the unique connection whose parallel transport satisfies the equation $\psi=\phi\circ T\circ \phi^{-1}.$

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  • $\begingroup$ thank you for your answer.I ,ll fix the typos in my post as you pointed out. $\endgroup$
    – Ali Taghavi
    Commented Jul 4, 2016 at 11:37
  • $\begingroup$ May ellaborate why this connection work or please give a reference.moreover what about the Riemannian part of my question?Does your connection automatically work? $\endgroup$
    – Ali Taghavi
    Commented Jul 4, 2016 at 11:40

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