Let $G$ be a connected Lie group and let $\xi_i$, $i=1,...,n$ be a basis of its Lie algebra (say, of left-invariant vector fields). We let $B_i$, $i=1,...,n$ be given symmetric sections of $TG \otimes TG$. I ask if there exists a symmetric section $A$ of $TG \otimes TG$ such that $\mathcal L(\xi_i)A=B_i$ for each $i$.
One integrability condition may be written down immediately. Let $f_{ij}^k$ be the structure constants, so that $[\xi_i, \xi_j]=\sum_k f_{ij}^k \xi_k$. Equation I'm after implies a cocycle condition $\mathcal L(\xi_i) B_j - \mathcal L(\xi_j) B_i = \sum_k f_{ij}^k B_k$ (in fact I'm asking if the Lie algebra cocycle $B_i$ is a coboundary). Let's assume that $B_i$ indeed satisfies this condition.
I am interested both in a local solution (say, in a neighbourhood of the neutral element) as well as global aspects of this problem.
Let $\chi_i$ be a basis of right-invariant vector fields on $G$, so that $[\xi_i, \chi_j]=0$ and $[\chi_i, \chi_j] = \sum_k f_{ij}^k \chi_k$. Decompose $A = \sum_{k,l} a^{kl} \chi_k \otimes \chi_l$, $B_i = \sum_{k,l} b_i^{kl} \chi_k \otimes \chi_l$. Equation $\mathcal L(\xi_i) A = B_i$ is then equivalent to $\xi_i(a^{kl})=b_i^{kl}$. Integrability condition given in the question and the Poincare lemma yield local solvability, while global obstructions to global solvability are seen to lie in the first de Rham cohomology group.