all its components have the same homotopy type. When we need to choose a component, we often take the connected component which includes the identity map. We denote this component by $\operatorname{Aut}^I _H (F_H)$.

Since $\operatorname{Aut}_H (F_H)$ is a monoid, the usual construction of classifying spaces for monoids applies, and we get a universal fibration $E\operatorname{Aut}_H (F_H) \to B\operatorname{Aut}_H (F_H)$ with fiber $\operatorname{Aut}_H (F_H)$. From this one also obtains a fibration $$F_H \to E_H \to B\operatorname{Aut}_H (F_H)$$ where $E_H =E\operatorname{Aut}_H (F_H) \times _{\operatorname{Aut}_H (F_H)} F_H$. This is actually an $H$-fibration with trivial $H$-action on the base space. It turns out that this fibration is a universal fibration for all $H$-equivariant fibrations with trivial action on the base space.

\[thm:classification\] Let $H$ be a finite group, $F_H$ be a finite $H$-CW-complex, and $B$ be a CW-complex with trivial $H$-action. Then, there is a one-to-one correspondence between $H$-fiber homotopy classes of $H$-fibrations over $B$ with fiber $F_H$ and the set of homotopy classes of maps $B \to B\operatorname{Aut}_H (F_H )$. The correspondence is given by taking the pullback of the universal $H$-fibration described above via the map $f:B \to B\operatorname{Aut}_H(F_H)$.

This theorem is proved in [@guclukanthesis] in full detail. The proof is 