The transition functions of a vector bundle over the field $\mathbb{F}$ are in $GL(n,\mathbb{F})$. Such a vector bundle has structure group equal to a a subgroup of $GL(n,\mathbb{F})$.
Do all fibre bundles have a structure group? If the transition functions of a vector bundle don't form a subgroup, does it make sense to take the "smallest" subgroup that contains all transition functions as the structure group?
Let $F \to E \xrightarrow{\pi} B$ be a continuous fibre bundle and let $\{U_i\}$ be an open cover of $B$ such that there are homeomorphisms $\varphi_i : \pi^{-1}(U_i) \to U_i\times F$ satisfying $\operatorname{pr}_1\circ\varphi_i = \pi|_{\pi^{-1}(U_i)}$; we call the collection $\{(U_i, \varphi_i)\}$ a local trivialisation. Then $\varphi_i\circ\varphi_j^{-1} : (U_i\cap U_j)\times F \to (U_i\cap U_j)\times F$ is given by $(x, \xi) \mapsto (x, t_{ij}(x)\xi)$ where $t_{ij} : U_i\cap U_j \to \operatorname{Homeo}(F)$ is continuous.
Let $G$ be a topological group which acts continuously and effectively on $F$. This is equivalent to $G$ being (isomorphic to) a subgroup of $\operatorname{Homeo}(F)$. We say the fibre bundle $F \to E \xrightarrow{\pi} B$ has structure group $G$ if there exists a trivialising open cover such that for all $i, j$ with $U_i\cap U_j \neq \emptyset$, $t_{ij} : U_i\cap U_j \to G \subseteq \operatorname{Homeo}(F)$. In particular, any fibre bundle with fibre $F$ has structure group $\operatorname{Homeo}(F)$.
The same discussion for smooth fibre bundles shows that any smooth fibre bundle with fibre $F$ has structure group $\operatorname{Diff}(F)$.
