Let $p:E\longrightarrow B$ be a smooth submersion and let $\sigma:p^*(TB)\longrightarrow TE$ be a complete connection with respect to $p$.
Given a path $\gamma:I\longrightarrow B$ we have the notion of holonomy of $\sigma$ along $\gamma$ which is a diffeomorphism $$\textrm{Hol}^\sigma_{\gamma}: E_{\gamma(0)}\longrightarrow E_{\gamma(1)},$$ where $E_{\gamma(j)}:=p^{-1}(\gamma(j))$. I could have choosen another complete connection $\sigma^\prime$ to get $\textrm{Hol}^{\sigma^\prime}_\gamma$.
Now, let us fix $e\in E_{\gamma(0)}$. How can I find a path on $E_{\gamma(1)}$ connecting $\textrm{Hol}^\sigma_{\gamma}(e)$ and $\textrm{Hol}^{\sigma^\prime}_\gamma(e)$?
A fact that I know is that $\sigma$ and $\sigma^\prime$ differ by a vertical $1$-form on the basis. More precisely, $\sigma$ and $\sigma^\prime$ define a map $$\theta^{\sigma_1, \sigma_2}\in \textrm{Hom}_{C^\infty(B)}(\Gamma(TB), \Gamma(VE)),$$ where $VE$ is the vertical bundle of $E$. Explicitly, $\theta^{\sigma, \sigma^\prime}$ is given by $$\theta^{\sigma, \sigma^\prime}(X)_x:=(\sigma_x-\sigma^\prime_x)(X_{p(x)})\quad \quad (x\in E)$$ where $\sigma_x:=(dp_x|_{H_x})^{-1}:H_x\longrightarrow T_{p(x)} B$, $\sigma_{x}^\prime:=(dp_x|_{H_x^\prime})^{-1}: H_x^\prime\longrightarrow T_{p(x)} B$. Here $H_x$ and $H_x^\prime$ are the horizontal subspaces of $T_x E$ determined by $\sigma$ and $\sigma^\prime$, respectively.
Thanks.
If the fibre of $p(Hol_{\gamma}^{\sigma}(e))=p(Hol_{\gamma}^{\sigma'}(e))$ is connected and is a manifold, there will be a path between $Hol_{\gamma}^{\sigma}(e))$ and $Hol_{\gamma}^{\sigma'}(e))$ since a connected manifold is connected by arcs.