$\mathcal{T}$ is not a Lie group when $n>1$.
You did not say whether you wanted $\mathcal{T}$ to be all possible sequences of compositions of these generating sets, but, if you did, then it is clear that $T$ is not a Lie group, in the sense that it is not defined as the set of solutions of some system of PDE for transformations of $\mathbb{R}^n$. For one thing, the group that they generate would properly contain the conformal group acting on $S^n$, which is known to be a maximal Lie group, i.e., there is no group (in Lie's sense) between the conformal group and the full diffeomorphism group. (NB: The group of analytic diffeomorphisms of $S^n$ is not a subgroup of the full diffeomorphims in Lie's sense because it is not defined as the set of solutions of some system of PDE.)
In particular, no group that contains $\mathcal{T}$ can preserve any geometric structures of the kind you mention because this would define a PDE that $\mathcal{T}$ satisfies.