Do fiber and structure group determine the fiber bundle over a given space? So, let's say $B$ is a honest topological space (path connected and locally simply-connected), and we are given two fiber bundles over $B$
$$ F_i \hookrightarrow E_i \to B, \qquad i=1,2 $$
with structure groups $G_i$. Now suppose that


*

*the fibers are the same: $F_1=F_2$.

*the structure groups are the same: $G_1 = G_2$.
Can we conclude that the total spaces are also the same, i.e. that $E_1=E_2$ ?  
 A: Definitely not.
Here's a simple example. Depending on the choice of transition functions, a bundle of the form
$$[-1,1] \hookrightarrow E \to S^1$$
and structure group $\Bbb Z/2 = \{-1,1\}$ (acting on the fiber $[-1,1]$ via multiplication) can have total space a cylinder or a Möbius band.
Edit: Let $[-1,1] \hookrightarrow E \to S^1$ be a fiber bundle with structure group $\Bbb Z/2 = \{-1, 1\}$ as above. I'll construct the cases where $E$ is the cylinder $S^1 \times [-1,1]$ and where $E$ is the Möbius band using transition functions.
Suppose that $E$ is trivialized over $U_1 = (-\varepsilon, \pi +\varepsilon)$ and $U_2 = (\pi-\varepsilon, \varepsilon)$ (here $\varepsilon$ is just a small positive number and we are thinking of $S^1$ as $\Bbb R/2\pi\Bbb Z$). We have that
$$U_1 \cap U_2 = (-\varepsilon, \varepsilon) \cup (\pi-\varepsilon, \pi+\varepsilon).$$
Assume that on the component $(\pi-\varepsilon, \pi+\varepsilon)$ of $U_1 \cap U_2$, the transition function is given by
$$\theta_{12}(x) = 1 \text{ for all } x \in (\pi-\varepsilon, \pi+\varepsilon).$$
Since a transition function must be continuous and $\Bbb Z/2$ is disconnected, we see that the only possibilities for $\theta_{12}|_{(-\varepsilon,\varepsilon)}$ are
\begin{align*}
 \text{(a)} & \quad \theta_{12}(x) = 1 \text{ for all } x \in (-\varepsilon,\varepsilon), \\
 \text{(b)} & \quad \theta_{12}(x) = -1 \text{ for all } x \in (-\varepsilon,\varepsilon).
\end{align*}
Case (a) gives the trivial bundle over $S^1$. In case (b), we have that the total space $E$ is the Möbius band.
To visualize the two cases (a) and (b) above, note that based on our choices so far, we have a strip $[-\varepsilon,2\pi+\varepsilon] \times [-1,1]$ and we need to glue the two ends $[-\varepsilon,0] \times [-1,1]$ and $[2\pi, 2\pi+\varepsilon] \times [-1,1]$ together to obtain $E$. Choice (a) just identifies the two ends trivially and gives us the cylinder $S^1 \times [-1,1]$. Choice (b) makes us rotate one of the ends before gluing, introducing a twist and hence giving us a Möbius band.
If we fix the base space $B$, fiber $F$, structure group $G$, and transition functions $\{\theta_{\beta\alpha}\}$, then the total space $E$ is determined. Essentially, $E$ is made up of pieces $U_\alpha \times F$ where $\{U_\alpha\}$ is an open cover of $B$, and the transition functions $\{\theta_{\beta\alpha}\}$ tell us how to glue together these pieces along overlaps where $U_\alpha \cap U_\beta \neq \varnothing$. The transition functions are just a recipe for gluing together the locally trivial pieces of the bundle to obtain $E$.
A: Consider your question for principal $G$-bundles over an honest space $M$. Note that the structure group of a principal $G$-bundle is $G$. 
Isomorphism classes of principal $G$-bundles over $M$ are in one-to-one correspondence with homotopy classes of maps $M\to BG$, the classifying space of $G$. So in particular, the fiber $G$ and base $M$ characterize the bundle if and only if $[M,BG]$ is trivial.
See also this MSE question: Classification of general fibre bundles
