I want to let $E$ be a rank $n$ complex vector bundle on a smooth manifold $X$. A connection $\nabla$ gives us a way of differentiating local or global sections of $E$. We define it to be a $\mathbb{C}$-linear sheaf morphism
$$\nabla : \Omega^{0}(E) \longrightarrow \Omega^{1}(E),$$
satisfying the Leibniz rule $\nabla(f \cdot s) = f \cdot \nabla(s) + s \cdot df$ for all local section $s$ and local complex functions $f$. A connection $\nabla$ induces a connection (which we also call $\nabla$) on $\text{det}(E) = \bigwedge^{n}E$
$$\nabla: \Omega^{0}(\text{det}(E)) \longrightarrow \Omega^{1}(\text{det}(E))$$
Now, assume $E$ has a trivial determinant; in other words $\text{det}(E)$ is isomorphic to the trivial line bundle on $X$. Now my question is, does it follow from what I said above that there exists sections $s_{1}, \ldots, s_{n}$ of $E$ such that $\nabla(s_{1} \wedge \ldots \wedge s_{n})=0$, or is this rather a condition we should impose in the definition of a "compatible connection" with the triviality of $\text{det}(E)$?
I'm leaning towards this being an additional property one may ask for. If you look at the Leibniz rule, this seems equivalent to asking the induced connection on $\text{det}(E)$ to coincide with the exterior derivative. In other words, we are asking that $\nabla$ induce the trivial connection on the trivial bundle $\text{det}(E)$.
EDIT: My motivation for this question comes from the book "Instantons and Four-Manifolds" by Freed and Uhlenbeck. On Page 30 (in the second edition) they are briefly introducing the standard notion of a connection on a complex vector bundle. They mention that if the bundle has a hermitian metric, you would ask a connection to additionally satisfy the usual compatibility condition. If the metric is Hermitian, this breaks the structure group down from $GL_{n}(\mathbb{C})$ to $U(n)$. To get to $SU(n)$, you also need the determinant to be trivial. So...they write:
"If, in addition, the group of the bundle is $SU(n)$, then we require the existence of orthonormal local sections $\sigma_{1}, \ldots, \sigma_{n}$ for which
$$\nabla(\sigma_{1} \wedge \ldots \wedge \sigma_{n})=0"$$