Do we have an equivariant version of integrability theorem of flat connections?

Question

I am reading Donaldson and Kronheimer's book The Geometry of Four-Manifolds. In page 48, I found Theorem 2.2.1:

Let $H$ be the hypercube $H=\{\mathbf{x}\in \mathbb{R}^d|~|x_i|<1\}$. If $E$ is a bundle over $H$ and $A$ is a flat connection on $E$. There is a bundle isomorphism taking $E$ to a trivial bundle over $H$ and $A$ to the product connection.

Now I am looking for an equivariant version of the above theorem. In more details, let $B$ be the ball $B=\{\mathbf{x}\in \mathbb{R}^d|~||\mathbf{x}||<1\}$. Let $G\subset SO(d)$ be a finite subgroup. Then $G$ acts on $B$. Let $E$ be an $G$-equivariant bundle over $B$ and $A$ is a $G$-invariant flat connection on $E$. Then do we have a $G$-equivariant isomorphism taking $E$ to a $G$-equivariant trivial bundle and $A$ to the product connection?

By $G$-equivariant trivial bundle I mean a bundle of the form $B\times V$ on which $G$ acts diagonally and the action on $V$ is linear.

I have this question because the proof of Theorem 2.2.1 in Donaldson and Kronheimer's book involves ODE's of each coordinate hence is not $G$-invariant itself.

Answer 1

The answer is "yes". The trivialisation is given by parallel transport along radial lines starting at $0\in B$. If the connection is $G$ equivariant, then elements of $G$ map parallel sections to parallel sections, so you get the desired "$G$-equivariant trivial bundle".