Let $C^{\infty,\star}(\mathbb{R}^d)$ denote the non-commutative topological group of smooth diffeomorphisms from $\mathbb{R}^d$ to itself with $\circ$ as multiplication and let $\emptyset\subset X\subseteq C^{\infty,\star}(\mathbb{R}^d)$. Are there (non-trivial) conditions one can use to verify if the closure of the group generated by $X$, $\overline{\langle X\rangle}$, equals to all of $C^{\infty,\star}(\mathbb{R}^d)$?

I'm thinking of some type of a group-theoretic version of the Weirestrass's result (since obviously these objects need not be algebras).

Let $C^{\infty,\star}(\mathbb{R}^d)$ denote the non-commutative topological group of smooth diffeomorphisms from $\mathbb{R}^d$ to itself with $\circ$ as multiplication and let $\emptyset\subset X\subseteq C^{\infty,\star}(\mathbb{R}^d)$. Are there (non-trivial) conditions one can use to verify if the closure of the group generated by $X$, $\overline{\langle X\rangle}$, equals to all of $C^{\infty,\star}(\mathbb{R}^d)$?

I'm thinking of some type of a group-theoretic version of the Weierstrass's result (since obviously these objects need not be algebras).