Group action by a semi-direct product. Let $G$ be a group and $H,N$ subgroups, of which $N$ is a normal subgroup. 
Suppose that $G= H \ltimes N$ and that $H \cap N = 0$. 
Is any action of $G$ on a set $X$ equivalent to an action of $N$ on $X$ followed by an action of $H$ on $X$? 
 A: It might depend on what you mean by equivalent here. My interpretation is that $\{\psi:G\to Sym(X)\}$ is in bijection with $\{\psi:N\to Sym(X)\}\times\{\psi:H\to Sym(X)\}$ which is not true.

There is an injection, but not a bijection:

If $K$ is a subgroup of $G$ then any action of $G$ on $X$ restricts to an action of $K$ on $X$, so we do get the injection $\psi\mapsto(\psi|_N,\psi|_H)$.
However a pair of actions $(\psi_N,\psi_H)$ only induce an action of $G$ if for all $h\in H$ and $n\in N$ we have $\psi_N(n^h)=\psi_N(n)^{\psi_H(h)}$.
There is an example where this does not hold with $|G|=6$ and $|X|=4$. 
A: If any $g\in G$ can be written an $g=hn$ then the action of $g$ is a action of an element in $N$ followed be an action of an element in $H$.
In but this is not "equivalent to an action of $N$ on $X$ followed by an action of $H$ on $X$".
Suppose $X=G$ and the action of $N$ is right action$n.x=xn^{-1}$ and the action of $H$ is the left action $h.x=hx$.
Then this does not induce a action on $G$ (in the case $N$ and $H$ do not commute), since if $g_1=h_1n_1$ and $g_2=h_2n_1$ we get that
$$g_2(g_1.x)=g_2(h_1xn_1^{-1}=h_2h_1xn_1^{-1}n_2^{-1}=(h_2h_1)x(n_2n_1)^{-1}$$
but $g_2g_1=h_2n_2h_1n_1\ne h_2 h_1 n_2n_1$ for $h_1$ which does not commute with $n_2$. Hence $(g_2g_1).x\ne g_2(g_1.x)$.
