Are vector bundles given by their monodromy?

It's easy to show that there's a bijection $$\{ \text{vector bundles }E \to X \text{ with a flat connection }\nabla\} \longleftrightarrow \{\text{hom's }\pi_1(X,x)\to \text{GL}(V)\}$$

But removing the flatness condition, is it true that a vector bundle $E\to X$ with a connection $\nabla$ is uniquely specified by the map $$\{\text{loops at }x\} \longrightarrow \text{GL}(V) \ \ \ ?$$ (This map certainly determines, for instance, the curvature of $\nabla$, and my intuition says that a vector bundle is probably specified by ''local monodromy'' (monodromy around small contractible loops i.e. its curvature) and ''global monodromy'' (monodromy around a set of representatives of $\pi_1(X,x)$))

Second, can we say anything about which of these maps arise from a vector bundle with connection?

Here is the answer.

Theorem. Conjugacy classes of homomorphisms $\Phi X\to G$ correspond to $G$ bundles with connections on $X$.

Proof sketch: Given a connection $\nabla$, parallel transport gives the homomorphism (up to conjugation). Conversely, a homomorphism allows us to turn $\Phi X$ bundles into $G$ bundles, so we get $G$ bundle $(PX\times G)/\Phi X$. To give a connection it is enough to know how to horizontally lift curves. Take a curve $\gamma(t)$ to $(p(t),g)$, where $g$ is determined uniquely by the choice of lift of $\gamma(0)$ and $p(t)$ is any path from $x$ to $\gamma(t)$ (which varies nicely with $t$). $\ \ \ \square$

Notation: $G$ is a Lie group, $X$ a manifold and $\Phi X$ its smooth loop space: the piecewise smooth basepoint preserving maps $S^1\to (X,x)$, up to orientation preserving reparametrisations of $S^1$. $\Phi X$ is a topological group. Similarly, write $PX$ for the smooth path space of $(X,x)$; $PX\to X$ is a $\Phi X$ bundle.