Let $M$ be a manifold and $(P\to E\overset\pi\to M, \omega: TE\to\mathfrak g)$ a Cartan geometry on $M$. I have seen the following statement:
Let $f:U\to V$ be a local automorphism where $U,V$ are connected open subsets of $M$. $f$ is uniquely determined by the value of its lift on an arbitrary point $\hat x\in \pi^{-1}(U)$.
In the case that the Cartan geometry encodes a Semi-Riemannian geometry the statement is clear to me, essentially being the statement that local isometries are uniquely determined by their value and their derivative at a single point. However I don't see the argument for an arbitrary Cartan geometry.
I know that lifts of automorphisms to functions on $E$ that leave $\omega$ invariant exist and are unique, but do not see the step giving the above Lemma.
As in the Semi-Riemannian case you mention, it is sufficient to prove that $f$ is uniquely determined on some neighborhood of a point $x\in U$ by the value of its lift in some point $\hat x$ lying over $x$. (From this the claim follows directly from connectedness.) To prove the latter statement, one uses the flow lines of "constant vector fields" instead of horizontal lifts of geodesics. These are the vector fields $\tilde X\in\mathfrak X(E)$ corresponding to $X\in\mathfrak g$ and defined by $\omega(\tilde X)=X$. Denoting by $F$ the lift of $f$, the fact that $f^*\omega=\omega$ easily implies that $F^*\tilde X=\tilde X$ for each $X\in\mathfrak g$. Denoting by $c_1(t)$ and $c_2(t)$ the flow linear of $\tilde X$ through $\hat x$ and $F(\hat x)$, respectively, this implies that $F(c_1(t))=c_2(t)$. Since this works for each $X$ and the resulting flow lines fill an open neighborhood of $\hat x$ in $E$, this completes the argument.