No, it is not always true. For instance, suppose $G$ is the forgetful functor from monoids to sets. Then $F$ is the free monoid functor, and the two maps in question are
$$((x_{1,1}, \ldots, x_{1,n_1}), \ldots, (x_{m,1}, \ldots, x_{m,n_m})) \mapsto (x_{1,1} \cdots x_{1,n_1}, \ldots, x_{m_1} \cdots x_{m,n_m})$$
$$((x_{1,1}, \ldots, x_{1,n_1}), \ldots, (x_{m,1}, \ldots, x_{m,n_m})) \mapsto (x_{1,1}, \ldots, x_{1,n_1}) \cdots (x_{m_1}, \ldots, x_{m,n_m})$$
More concretely, if we identify $F G 1$ with $\mathbb{N}$ and $F G F G 1$ with finite lists of natural numbers, then the two maps are (respectively)
$$(x_1, \ldots, x_m) \mapsto m$$
$$(x_1, \ldots, x_m) \mapsto x_1 + \cdots + x_m$$
which are obviously not the same.
It's worth pointing out that the equation holds if and only if $F$ and $G$ form an idempotent adjunction. In particular, it is true when $G$ is fully faithful.