Let $\mathcal{T}$ be a set of tensors in $\mathbb{R}^n$, and let $G_{\mathcal{T}}$ be a subgroup of $GL(n)$ defined as $$ G_{\mathcal{T}} = \{ g \in GL(n) \mid g \cdot T = T, \, \forall T \in \mathcal{T} \}, $$ where "$\cdot$" denotes the appropriate tensor transformation rule under the action of $g \in GL(n)$. I know $G_{\mathcal{T}}$ is a closed subgroup of $GL(n)$ (right?).
Now, given a closed subgroup $G$ of $GL(n)$, can we find a set of tensors $\mathcal{T}$ such that $G=G_{\mathcal{T}}$?
No, let $G\neq GL_n(\mathbb{R})$ be a closed subgroup which contains $\lambda I_n$ for some $|\lambda|> 1$, e.g. $\mathbb{Z}^n$ or diagonal matrices. Suppose $G=G_{\mathcal{T}}$. Then $(\lambda I_n)\cdot T =\lambda^k T$ for a $k$-tensor $T$, and so, $(\lambda I_n)\cdot T=T$ implies $T=0$. Therefore, since $G\subseteq G_{\mathcal{T}}$ it is $\mathcal{T}\subseteq\{0\}$ and $G_{\mathcal{T}}=GL_n(\mathbb{R})\neq G$, a contradiction.