Suppose $p_1:E_1\to B$ and $p_2: E_2\to B$ are two compact principal $\mathbb{T}^d$-bundles over the same base $B$. Suppose there exists a fibre-preserving map $F:E_1\to E_2$ that covers the identity $\text{id}_B$, i.e., $p_2\circ F=\text{id}_b\circ p_1$.
By compactness then I think I could conclude that $F$ induces the same automorphism on the the first homology (and the fundamental group) of every $\mathbb{T}^d$ fibre, i.e., $F_b:p_1^{-1}b\to p_2^{-1}b$ on the first homology of $\mathbb{T}^d$ is the same matrix on $\mathbb{Z}^d\to\mathbb{Z}^d$ at every $b$.
Now I want to further assume that $(F_b)_*:\mathbb{Z}^d\to\mathbb{Z}^d$ is the identity.
Then I wonder if I could kind of "tighten this map up". More precisely, I wonder if I could find a map $G\simeq F$ such that $G(t.e)=t.G(e)$ where $t\in\mathbb{T}^d,\ e\in E_1$. That is, $F$ is homotopic to a map that is identity in the fibres. This in particular implies that $E_1$ and $E_2$ are isomorphic.
I think this might rely on the fact that $\mathbb{T}^d$ is a $K(\pi,1)$. I was trying to chase through long exact sequence but wasn't very successful.
Thanks in advance for any help!