Finding Bijection between subgroups of group $G = NU$. 
Let $N \unlhd G$ and $U \le G$ with $G = NU$. Then there exists an inclusion-preserving bijection from the set of all subgroups $X$ with $U \le X \le G$ on the set of all $U$-invariant subgroups $Y$ with $U \cap N \le Y \le N$.

A subgroup $V \le G$ is called $U$-invariant for some subgroup $U \le G$, if $V^g = V$ for each $g \in U$.
It guess the map might be $\varphi(X) = X \cap N$, I could show injectivity by using the so called Dedekind identity (*) by which if $X \cap N = Y \cap N$ and factoring $X = U(X\cap N), Y = U(Y\cap N)$ we have $$ X = U(X\cap N) = U(Y\cap N) = Y. $$ 
But how to show surjectivity?
(*) The Dedekind identity as I know it goes like this, let $G = UV$ for two subgroups $U,V \le G$. Then every subgroup $H$ with $U \le H \le G$ has a factorisation $H = U(V\cap H)$.
 A: Here's a detailed version, please see if it would be a helpful one.

Proof. 
For each $X$, $U$ is a subgroup of $X$, hence normalizes $X$; $N$ is a normal subgroup of $G$, so $U$ certainly normalizes $N$. Thus, $U$ normalizes $X\cap N$, i.e., $X\cap N$ is $U$-invariant. Now we can define a map
      \begin{align*}
 \varphi~:~\mathfrak{A}=\{ X\mid U\le X\le G \} &\longrightarrow \mathfrak{B}=\{Y\mid Y \mbox{ is a }U\mbox{-invariant subgroup}\}\\X&\longmapsto X\cap N.
 \end{align*}
  The map is well defined since $U\leq N_G(X\cap N)$. 
For the injectivity: $\forall\, X_1\cap N,X_2\cap N\in \mathfrak{B}$, if $X_1\cap N=X_2\cap N$, then by Dedekind Identity, $X_1=U(X_1\cap N)=U(X_2\cap N)=X_2$. 
The map is also surjective. $\forall\, Y\in \mathfrak{B}$, $u\in U$, since $Y$ is $U$-invariant, $Y^u=u^{-1}Yu=Y$, $Yu=uY$. Hence, $YU=UY$, $YU\leq G$. Note that $U\cap N\leq U$, $NU=G$, $(U\cap N)U\subseteq YU\subseteq NU\Longrightarrow U\leq YU\leq G$. So we can apply Dedekind Identity to $YU$, $YU=UY=U(UY\cap N)\Longrightarrow Y=UY\cap N$. Since $UY\in \mathfrak{A}$, each $Y$ has a corresponding $X\in \mathfrak{A}$, therefore $\varphi$ is surjective. 
The bijection is hence proved.

Note: It's fundamentally based on Nicky Hekster's hints. 
