A semidirect product of two groups $H$ and $N$ is the same as a group homomorphism $\varphi: H\rightarrow \mathrm{Aut}(N)$ i.e. an action of $H$ on $N$. Groupies usually drop the $\varphi$ from the notation and simply write $n^h$ instead of $n^{\varphi h}$. The resulting multiplication on $H\times N$ is then given by $(h,n)(k,m) = (hk,n^km)$.
Likewise a semidirect product of two monoids $H$ and $N$ is the same as a monoid homomorphism $\varphi: H\rightarrow \mathrm{End}(N)$. The multiplication is given in the same way.
Now consider a functor $F: C\rightarrow \mathrm{Cat}$ (I ignore dualization, so this correspond to an op-fibration) and suppose that
(1) $C$ is a monoid (i.e. has exactly one object $*$)
(2) The category $F(*)$ is also a monoid.
Because an endofunctor on $F(*)$ is the same as a monoid homomorphism, this is the same as giving a monoid homomorphism from $C$ to $\mathrm{End}(F(*))$.
In fact, if you go through the Grothendieck construction for the corresponding op-fibration, you will produce exactly the above multipilcation. The similarity is also visible in the general case if you view $F: C\rightarrow \mathrm{Cat}$ as a category action of $C$, drop the $F$ from the notation and write $uf$ instead of $F(f)u$ for $f:c\rightarrow c'$ and $u\in F(c)$:
For $c,c',c''\in C$, $x\in F(c)$, $x'\in F(c')$, $x''\in F(c'')$, $f:c\rightarrow c'$, $u:F(f)x\rightarrow x'$, $g:c'\rightarrow c''$, $u:F(g)x'\rightarrow x''$ consider the maps $(f,u):(c,x)\rightarrow (c',x')$ and $(g,v):(c',x')\rightarrow (c'',x'')$ in the Grothendieck construction. If we write composition of maps from left to right and apply the above mentioned notational simplification, then the composition of $(f,u)$ and $(g,v)$ can be written as $(fg,(ug)v)$ where $(xf)g$ is the composite $xfg \stackrel{ug}{\rightarrow} x'g \stackrel{v}{\rightarrow} x''$
There is also a very nice description in Section 12.2 ("The Grothendieck construction") of Barr/Wells: Category Theory for Computing Science.