Let $G$ be a (compact, semisimple) Lie group, $\rho\in\operatorname{Hom}\left(G,\operatorname{Aut}\left(V\right)\right)$ a non-trivial representation of $G$ on the (real) vector space $V$ (w.l.o.g. we can assume that $\rho$ is orthogonal). If $\varphi\in{C^{\infty}\left(M,V\right)}$, where $M$ is a smooth (connected and simply connected) manifold and $g\colon{M}\to{G}$ is a map such that: $$\varphi\left(x\right)=\rho\left(g\left(x\right)\right)u\quad,\quad\forall{x}\in{M}$$ for some $u\in{V}$, then is it implied that $g$ is smooth?
If not, can we deduce anything about the differentiability of $g$?
The difficulty here is that the map $g \mapsto \rho(g(x)) u$ is not injective. Therefore, the condition $\varphi(x) = \rho(g(x)) u$ does little to constrain $g$.
Even if you assume that $\rho$ is a faithful representation (i.e. $g \mapsto \rho(g) $ is injective), this still isn't enough to guarantee that the map $g \mapsto \rho(g(x)) u$ is injective. This is because different automorphisms of $V$ may act on $u$ in the same way.
For example, take $G = SO(3)$ and take $V = \mathbb R^3$ with $\rho$ being the fundamental representation of the group $G$. Pick $u = (1,0,0)$. Take $M = S^1 $ and define $\varphi(\theta) = (\cos \theta, \sin \theta, 0)$. Then, of course, we can pick $g(\theta)$ to be the element of $SO(3)$ corresponding to the rotation about the $z$-axis through an angle $\theta$. This choice of $g$ is smooth, and obeys $\varphi(\theta) = \rho(g(\theta))u$. But we could just as well pick $g(\theta)$ to be the rotation about the $z$-axis through an angle $\theta$, followed by a rotation about the axis $(\cos \theta, \sin \theta, 0)$ through any arbitrary angle; moreover, we can choose the arbitrary angle differently for each $\theta$. This choice of $g$ still obeys $\varphi(\theta) = \rho(g(\theta))u$, but it is not even continuous.