Suppose I have two closed, aspherical manifolds, $F$ and $B$, with $K = \pi_1(F)$ and $Q = \pi_1(B)$, and two group extensions $1 \to K \to G_1 \to Q \to 1$ and $1 \to K \to G_2 \to Q \to 1$ with $G_1$ and $G_2$ abstractly isomorphic but not congruent as group extensions. What can be said about 1) the characteristic classes of $F \hookrightarrow E_1 \twoheadrightarrow B$ and $F \hookrightarrow E_2 \twoheadrightarrow B$ and 2) whether $E_1$ and $E_2$ are homeomorphic as manifolds?
For instance, say I have fiber bundles $E_1^4$ and $E_2^4$ with base and fiber $T^2$ as follows.
Set $G_1 = \langle w_1, w_2, z_1, z_2\ |\ w_1w_2w_1^{-1}w_2^{-1}, w_1z_1w_1^{-1}z_1^{-1}, w_1z_2w_1^{-1}z_2^{-1}, w_2z_1w_2^{-1}z_1^{-1}, w_2z_2w_2^{-1}z_2^{-1}, (w_1^1w_2^4)^{-1}(z_1z_2z_1^{-1}z_2^{-1})\rangle$ and $G_2 = \langle u_1, u_2, z_1, z_2\ |\ u_1u_2u_1^{-1}u_2^{-1}, u_1z_1u_1^{-1}z_1^{-1}, u_1z_2u_1^{-1}z_2^{-1}, u_2z_1u_2^{-1}z_1^{-1}, u_2z_2u_2^{-1}z_2^{-1}, (u_1^2u_2^3)^{-1}(z_1z_2z_1^{-1}z_2^{-1})\rangle$.
Note $\begin{bmatrix} -2 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 4 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}$, so $G_1$ and $G_2$ should be isomorphic:
In particualr, define $\Phi: G_1 \to G_2$ by starting with $\Phi(z_j) = z_j$. Then, note $\begin{bmatrix} -2 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} -2 \\ -1 \end{bmatrix}$, so send $\Phi(w_1) = u_1^{-2}u_2^{-1}$. Next, $\begin{bmatrix} -2 & 1 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$, so let $\Phi(w_2) = u_1u_2$. This yields $\Phi(w_1w_2w_1^{-1}w_2^{-1}) = (u_1^{-2}u_2^{-1})(u_1u_2)(u_1^{-2}u_2^{-1})^{-1}(u_1u_2)^{-1} = e$. Similarly, $\Phi$ satifies the other slide relators involving the (u_i)'s and (z_j)'s. Finally, $\Phi(w_1w_2^4) = (u_1^{-2}u_2^{-1})(u_1u_2)^4 = (u_1^{-2}u_2^{-1})(u_1^4u_2^4) = u_1^2u_2^3$. So, $\Phi$ sends a relator of $G_1$ to a relator of $G_2$, and $\Phi$ is well-defined as a homomorphism of finitely-presented groups.
A similar argument, using the fact that $A = \begin{bmatrix} -2 & 1 \\ -1 & 1 \end{bmatrix} \in GL_2(\mathbb{Z})$, as $\det(A) = -1$, shows $\Phi^{-1}$ is a well-defined homomorphism. A third simple calculation shows $\Phi$ and $\Phi^{-1}$ are inverses of each other. Hence, $G_1$ and $G_2$ are isomorphic.
But, I don't think they're congruent group extensions.
An outline of how to produce a fiber bundle from a group extension of the fundamental groups of the fiber and base spaces may be found here http://deadbeatjeff.sdf.org/mathjax/fiberBundle.html
Now, if we write $e^0_{n_0}$ for the element of $C^0(T^2)$ that sends $z_0$ to $n_0z_0$, $e^1_{(n_1,n_2)}$ for the element of $C^1(T^2)$ that sends $(z_1,z_2)$ to $n_1z_1+n_2z_2$, and $e^2_{n_3}$ for the element of $C^2(T^2)$ that sends $z_3$ to $n_3z_3$, then is the characteristic class of $E_1$ is $\eta_1 = e^1_{(1,1)} \smile e^1_{(1,4)}$ while the characteristic class of $E_2$ is $\eta_2 = e^1_{(1,2)} \smile e^1_{(1,3)} $?
Freedman and Quinn proved that closed, aspherical 4-manifolds with isomorphic polycyclic fundamental groups (which $E_1$ and $E_2$ are) are homeomorphic.
So, I can't have the correct characteristic classes for the fiber bundles; what are the correct characteristic classes for the fiber bundles?