For $X$ a *paracompact* space, I am trying to classify all locally trivial fibration with base the suspension $SX = X \times [-1,1]\, /\, (X \times \{-1\} \cup X \times \{1\})$, and fiber-type a space $F$ *such that $G\_F = Homeo(F)$ with the C.O. topology is a topological group*. 


I duplicate the reasonning for classification of fibration with fiber-type $F$ over the sphere $S^n$. In that case, denoting the set of isomorphism classes of such fibrations by $[FB(S^n,F)]$, we get that
$$ [FB(S^n,F)] \simeq \pi\_{n-1}(G\_F) / \pi\_{0}(G\_F), $$
where $\pi\_{n-1}(G\_F)$ is the set of pointed homotopy classes $(S^{n−1}, ∗) → (G\_F , Id\_F )$, and the action of $\pi\_{0}(G\_F)$ on $\pi\_{n-1}(G\_F)$ is induced by the action of $G\_F$ on itself by conjugaison.


In the case of base $SX$, we have $SX = CX\_+ \cup CX\_-$ and $X \simeq CX\_+ \cap CX\_-$, with the cones $CX\_{\pm}$ being paracompact and contractile, so every fibration over $CX\_+$ or $CX\_-$ is trivial. If $\Phi\_{\pm}$ are the corresponding trivializations, $f\_0 \in G\_F$ a fixed homeomorphism of $F$, and $x\_0 \in X$ an arbitrary base-point, we can always choose $\Phi\_{\pm}$ s.t. $\Phi\_+(x\_0) = \Phi\_-(x\_0) = f\_0$, and so the transition function from $\Phi\_+$ to $\Phi\_-$ determine an application
$$\phi \colon (X,x\_0) \to (G\_F,Id\_F)$$
As in the case of $S^n$, the action of $G\_F$ on itself by conjugaison induces an action on the homotopy classes $[\phi]$ which depends only of the path-component of the element in $G\_F$. Lets denote the set of isomorphism classes of fibration with fiber $F$ by $[FB(SX,F)]$, and the set of equivalence classes of homotopy class under the action of $\pi\_{0}(G\_F)$ by 
$$[(X,x\_0),(G\_F,Id\_F)]\_\* = [(X,x\_0),(G\_F,Id\_F)]\,/\,\pi\_{0}(G\_F)$$
For $[\xi] \in [FB(SX,F)]$ and $[\phi]\_\* \in [(X,x\_0),(G\_F,Id\_F)]\_\*$, we can define an application
$$\theta \colon [FB(SX,F)] \to [(X,x\_0),(G\_F,Id\_F)]\_\*, \quad [\xi] \mapsto [\phi]\_\*$$
This is well-defined : choosing another set of trivializations $\Psi\_{\pm}$ gives a function $\psi$ s.t. $[\psi]\_\* = [\phi]\_\*$, changing the choice of the value at $x\_0$ from $f\_0$ to $g\_0$ also gives applications in the same equivalence class ; it is also easy to show that two isomorphic fibration $\xi$ and $\xi'$ gives function in the same equivalence class.


Now, $\theta$ is a bijection : it is surjective because from an application $\phi\colon (X,x\_0) \to (G\_F,Id\_F)$ we can reconstruct a fibration over $SX$ with $\phi$ as its transition function unique up to iso, an injective because if $\xi$ gives $\phi$ and $\xi'$ gives $\phi'$ and $[\phi]\_\* = [\phi']\_\*$, we can modify $\phi$ s.t. $\phi$ and $\phi'$ are homotopic by an homotopy $H$ and construct a fibration $\xi\_H$ over $SX \times [0,1]$, and $SX$ being paracompact we get
$$\xi \simeq \xi\_{\psi}|\_{SX \times {0}} \simeq \xi\_{\psi}|\_{SX \times {1}} \simeq \xi′$$
So we apparently get that
$$ [FB(SX,F)] \simeq [(X,x\_0),(G\_F,Id\_F)]\_\*$$


*But here $[FB(SX,F)]$ cannot depend of the choice of the base-point, but it seems to me that $[(X,x\_0),(G\_F,Id\_F)]\_\*$ does depend on that choice if $X$ is not a homogeneous space like $S^n$*.


So I think there must be a mistake somewhere, but where ??


Or contrary to the intuition, does $[(X,x\_0),(G\_F,Id\_F)]\_\*$ is independant of the choice of $x\_0$ ?



Locally trivial fibration over a suspension
It is independent of the choice of base point. 


Let $Map((X,x\_0), (G\_F,id))$ be the based mapping space (based at $x\_0$).
Let $Map(X, G\_F)$ be the free mapping space. Then we have a split short exact sequence of topological groups:


$$Map((X,x\_0), (G\_F,id)) \to Map(X, G\_F) \stackrel{ev\_{x\_0}}{\to} G\_F$$


(These are groups using pointwise multiplication). The last map, which evaluates at $x\_0$, is split by identifying $G\_F$ with the constant maps. Note that the inclusion of the constant maps doesn't depend on the choice of basepoint. 


In any case this induces a split short exact sequence on the corresponding groups of path components:
$$\pi\_0Map((X,x\_0), (G\_F,id)) \to \pi\_0Map(X, G\_F) \stackrel{ev\_{x\_0}}{\to} \pi\_0G\_F$$


Now the group structure on $\pi\_0Map((X,x\_0), (G\_F,id))$ might depend on identifying it as the kernel of the map $ev\_{x\_0}$, which of course depends on the base point. 


However we only care about it as a $\pi\_0G\_F$-set. From the above split exact sequence it follows that as sets with $\pi\_0G\_F$-action $[(X,x\_0), (G\_F,id)] = \pi\_0Map((X,x\_0), (G\_F,id))$ is isomorphic to the $\pi\_0G\_F$-set of, say, left cosets:
$$\pi\_0Map((X,x\_0), (G\_F,id)) \cong \pi\_0Map(X, G\_F)/ \pi\_0G\_F$$ 
The right-hand side is independent of the choice of basepoint $x\_0$, which tells us, perhaps surprisingly, that already $[(X,x\_0), (G\_F,id)]$ is independent of basepoint. 


It further follows that the quotient of this set by $\pi\_0G\_F$, which is your $[(X,x\_0), (G\_F,id)]\_\*$, is also independent of basepoint. 


