What is the difference between $A\rtimes B$ and $B\rtimes A$?
Could one group be normal in a group $G$ and at the same time not normal in another group?
The asymmetry of the $\rtimes$ symbol (with or without a specified action) suggests that $A$ and $B$ aren't quite interchangeable in the semi-direct product construction, so it is natural to wonder whether both $A\rtimes B$ and $B\rtimes A$ make sense for any given pair of groups $A$ and $B$ and, if so, whether there is any relationship between the two.
Expanding/correcting my comment...
Recall that the semidirect product $G \rtimes_\varphi H$ is defined via a homomorphism $\varphi: H \to \operatorname{Aut}(G)$. The underlying set of the semidirect product is cartesian product $G \times H$, and multiplication $*$ is defined by
$$(a,b)*(c,d)=(a\varphi_b(c),bd),$$ where $\varphi_b$ abbreviates the automorphism $\varphi(b)$. Different homomorphisms $\varphi$ may give non-isomorphic groups, so you need to specify your homomorphism. Just writing $G \rtimes H$ is ambiguous (although on occasion it might be an appropriate shorthand).
As Derek Holt points out in the comments, the semidirect product associated to the trivial homomorphism exists for any $G$ and $H$. By trivial, I mean $\epsilon:H \to \operatorname{Aut}(G)$ where $\epsilon_h:G\to G$ is the identity of $G$ for each $h \in H$. In this case we recover the direct sum $G \oplus H$:
$$(g_1,h_1) * (g_2,h_2) =(g_1\epsilon_{h_1}(g_2),h_1h_2)=(g_1g_2,h_1h_2).$$
So we can always arrange that $A\rtimes_\varphi B \simeq B \rtimes_\psi A$ (by taking trivial homomorphisms for both). For general $\varphi$ and $\psi$:
Both semidirect products contain $A$ and $B$ as subgroups: $A\simeq A \times \{e_B\} $, and $B\simeq \{e_A\} \times B$.
For $A$ and $B$ finite, $|A\rtimes_\varphi B|=|B\rtimes_\psi A|=|A|\cdot |B|$: this is the size of the underlying set.
Beyond this, structures of $A \rtimes_\phi B$ and $B\rtimes_\psi A$ depend on the given maps. The two needn't be isomorphic; in fact, $A \rtimes_\varphi B$ needn't be isomorphic to $B \rtimes_\alpha A$ for any choice of $\alpha: A \to \operatorname{Aut}(B)$.
For an example of this, let $A=\mathbb Z/7\mathbb Z$, $B=\mathbb Z/3\mathbb Z$. We can identify $B$ with the set $\{1,2,4\}$ under multiplication modulo $7$ (check this). Define $\varphi:B \to \operatorname{Aut}(A)$ by
$$\varphi_i([x])=[ix],$$ where $[y]$ is the class of an integer $y$ in $\mathbb Z / 7 \mathbb Z$ (verify this is well-defined). This group is non-abelian:
$$([3],2)*([2],1)=([3]\cdot \varphi_2([2]),2)=([3]\cdot[4],2)=([5],2),$$
while
$$([2],1)*([3],2)=([2]\cdot \varphi_1([3]),2)=([2]\cdot[3],2)=([6],2).$$
So we see that $a*b \neq b*a$ for some $a,b \in A \rtimes_\varphi B$. On the other hand, $|\operatorname{Aut}(B)|=2$; by order considerations, the only homomorphism $\alpha:A \to \operatorname{Aut}(B)$ is trivial. In particular, $B \rtimes_\alpha A \simeq B \oplus A$ is abelian, and not isomorphic to $A \rtimes_\varphi B$.
I take your second question to be about whether $B$ can be normal in $A \rtimes_\varphi B$, as well as $B\rtimes_\psi A$. $B$ is always normal in the latter, but "rarely" in the former. You can verify that $$(a,b)^{-1}=(\varphi_{b^{-1}}(a^{-1}),b^{-1}),$$ and that
$$(a,b)*(c,d)*(a,b)^{-1}=(a\varphi_{b}(c)\varphi_{bdb^{-1}}(a^{-1}),bdb^{-1}).$$
This implies that $ A \times \{e_B\} \rhd A \rtimes_\varphi B$: if $d=e_B$, the right hand side is in $A \times \{e_B\}$ for any $(a,b)$. On the other hand, $\{e_A\}\times B$ isn't necessarily stable under conjugation:
$$(a,b)*(e_A,d)*(a,b)^{-1}=(a\varphi_{b}(e_A)\varphi_{bdb^{-1}}(a^{-1}),bdb^{-1})=(a\varphi_{bdb^{-1}}(a^{-1}),bdb^{-1}).$$
This element lies in $\{e_A\} \times B$ if and only if $\varphi_{bdb^{-1}}(a^{-1})=a^{-1}$. Since this must hold for any choice of $a,b,d$, this implies $\varphi_{x}=\mathbb 1_A$ for any $x\in B$. Thus, $B \lhd A \rtimes_\varphi B$ if and only if $\varphi$ is trivial.