The Lie derivative of some tensor $\overline{U}$ is invariant with respect to a vector field $V$ if$$\mathcal{L}_\overline{V} (\overline{U}) = 0,$$ where $\mathcal{L}$ is the Lie derivative.
There is, at the same time, another sense of invariance. Suppose we have a Lie group $G$, and let $L_g$ denote left translation by $g\in G$. A vector field $\overline{U}$ on $G$ is "left-invariant" if $L_g: \overline{U}(e)\rightarrow\overline{U}(g) $ for all $g$, where $e$ is the identity.
How are these two definitions of invariance related? For example, suppose our group $G$ is $\mathbb{R}^1$ under addition. Then $L_g(x)=g+x$, and we could let $$\overline{V}=\left(L_g(x)-x\right)\frac{\partial}{\partial x}.\qquad(*)$$ If $\overline{U}$ is left-invariant, then $\mathcal{L}_\overline{V} (\overline{U}) = 0$ as desired. The relationship $(*)$, I think, could be extended for any linear Lie group. But in general, how do we find $\overline{V}$ such that $\mathcal{L}_\overline{V} (\overline{U}) = 0$ whenever $\overline{U}$ is left-invariant.
When the dimension is greater than $1$, you should not expect $\mathcal L_V U = 0$ for any single $V$ to tell you that $U$ is left-invariant, since left-invariance is about being symmetric in all possible directions, not just along a single flow. The following fact is the closest I can think of to what you're describing:
This is because the flow of a right-invariant field $V$ is $g \mapsto \exp(t V_e)g$, so $\mathcal L_V U = 0$ tells us that $U$ is invariant under left-multiplication by some neighbourhood of the identity, which generates the whole group.