= Solution
Use the <Hamiltonian vector field> convention $\iota_{X_t}\omega=dH_t$, which is forced by the sign of the formula in this part. This is the opposite of the default sign in the linked general article. Since $d\lambda=-\omega$, <Cartan's magic formula> gives
$$
\mathcal L_{X_t}\lambda
=d(\iota_{X_t}\lambda)+\iota_{X_t}d\lambda
=d(\iota_{X_t}\lambda-H_t).
$$
Differentiate the <pullback of a differential form> along the <smooth isotopy> and integrate from $0$ to $t$:
$$
\begin{aligned}
\rho_t^*\lambda-\lambda
&=\int_0^t\rho_s^*\mathcal L_{X_s}\lambda\,ds\\
&=d\int_0^t(\iota_{X_s}\lambda-H_s)\circ\rho_s\,ds.
\end{aligned}
$$
Thus the required <exact differential form> has the explicit primitive
$$
\boxed{F_t=\int_0^t(\iota_{X_s}\lambda-H_s)\circ\rho_s\,ds,\qquad\rho_t^*\lambda-\lambda=dF_t.}
$$
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