In Sharpe's text on Cartan geometry, he explains in section 1.5 on page 52 how to define an exterior derivative for maps into a parallelizable manifold $N$. Let $f: M \to N$ be a smooth map, and $\theta: TN \to V$ a fibrewise linear isomorphism. This is obviously the same as a trivialization of the tangent bundle $TN \cong N \times V$. Now we can define an exterior derivative by $$df := \theta \circ Tf.$$
For example, this is precisely how the Darboux derivative is defined if $N =G$ is a Lie group and $\theta: TG \to \mathfrak{g}$ is the Maurer-Cartan form (this generalizes and includes the case $G = \mathbb{R}$ for which we get the usual exterior derivative).
In general, the exterior derivative depends on the chosen trivialization $\theta: TN \to V$. For instance, given a map $\tau: N \to \text{GL}(V)$, we have another trivialization $\tau . \theta: TN \to V$ obtained by twisting by $\tau$. In general $d f = \theta \circ Tf$ will give a different result than $$d_\tau f := (\tau . \theta) \circ Tf.$$ For instance, given $v_x \in TM$ we have $$(d f)(v_x) := \theta(Tf(v_x)),$$ $$(d_\tau f)(v_x) := \tau(f(x)) . \theta(Tf(v_x)).$$ Now this all makes sense to me, but then Sharpe says that
"the exterior derivative computed relative to the trivialization $\theta$ is the same as that computed relative to the trivialization $\tau.\theta$ if and only if $\tau$ is constant."
Maybe I'm missing something obvious, but I don't see how we could have $d_\tau f= d f$ unless $\tau(x) = \text{id}_V$.
I have a feeling that this is an error, but I am not sure what Sharpe meant to write. I haven't been able to find errata for the text. I showed this to a few other grad students and they weren't sure what to make of it either.