A connection on $Hom( E,E)$ whose parallel transport is compatible to parallel transport of $E$ 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 
 A: 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 $\psi=\phi\circ T\circ \phi^{-1}.$
