To shed light on some questions:
The statement "$G=T\rtimes Q \Leftrightarrow Q \cong Aut(\mathcal{L}^n)$" is wrong.
The statement "if $Q \cong Aut(\mathcal{L}^n)$ and $T \cong \mathcal{L}^n$, then $G=T\rtimes Q$" is wrong.
The following is true:
Theorem: There is a 1-1 correspondence between symmorphic space groups of dimension $n$ and the conjugacy classes of finite subgroups of $GL_n(\mathbb{Z})$.
Explanations:
In the following I write $L$ instead of $\mathcal{L}^n$ for a lattice and $Aut_O(L) := Aut(L) \cap O_n$ instead of $Aut(\mathcal{L}^n)$, where $O_n$ is the orthogonal group and $Aut(L)$ is the group of automorphisms of $L$ in the group theoretic sense.
1) Choose $L := \mathbb{Z}^n$. Since columns of matrices in $O_n$ have Euklid norm 1, $Aut_O(L) =GL_n(\mathbb{Z})\cap O_n$ consists of the signed permutation matrices (there are $2^n \cdot n!$). For a subgroup $Q \le Aut_O(L)$ let $G_Q := \langle t_a, r_B \mid a \in L, B \in Q \rangle$, where $t_a$ is translation by $a$ and $r_B: \mathbb{R}^n \to \mathbb{R}^n, x \mapsto Bx$. Then $G_Q$ is the semi-direct product of its translational subgroup $T = \langle t_a \mid a \in L \rangle$ and of $\langle r_B \mid B \in Q \rangle \cong Q$. This shows that $(\Rightarrow)$ is false (just take any $Q \lvertneqq Aut_O(L)$).
In order to show that $(\Leftarrow)$ is also false, it suffices to find a counterexample in dimension 2 (I guess there are counterexamples in each dimension, but that would be a - nontrivial - topic for itself). $Aut_O(\mathbb{Z}^2)$ is the dihedral group $D_8$ of order 8 and the existence of such a counterexample follows from $H^2(D_8;\mathbb{Z}^2) = \mathbb{Z}/2$.
(Under the 17 space groups of dimension 2, there are exactly 2 with point group $D_8$ and by looking into the corresponding tables you should be able to figure out the non-split one to get an explicit presentation.)
2) This was just shown by the counterexample. You misread the explication in the Havana paper: The author merely shows the existence of a space group those point group is isomorphic to $Aut_O(L)$ - namely the semi-direct product (but he does not say that, if the point group is isomorphic to $Aut_O(L)$, then the space group is symmorphic!). This is to justify the subsequent definition of Bravais groups (Def. 42).
3) In the following write $L_0 := \mathbb{Z}^n$ and $T_0 = \langle t_a \mid a \in \mathbb{Z}^n \rangle$.
Let $\mathcal{C}_n$ be the set of conjugacy classes of finite subgroups
of $GL_n(\mathbb{Z})$ and let $\mathcal{I}_n$ be the set of isomorphism
classes of symmorphic space groups. Define a mapping
$$f: \mathcal{C}_n \to \mathcal{I}_n, [Q] \mapsto [G_Q]$$
where $[.]$ denotes the corresponding class on either side and $G_Q := L_0 \rtimes Q$, where $Q$ acts on $L_0$ through matrix multiplication.
Step 0: $f$ is well-defined
Let $Q,P \le GL_n(\mathbb{Z})$ be conjugated in $GL_n(\mathbb{Z})$,
i.e. there is $A \in GL_n(\mathbb{Z})$ such that $P = AQA^{-1}$. Using
$$\alpha: L_0 \to L_0, x \mapsto Ax,$$
$$\beta: Q \to P, B \mapsto ABA^{-1},$$ one immediately sees
$\beta(B)\cdot \alpha(x) = ABA^{-1}Ax = \alpha(Bx)$. Consequently
$$L_0 \rtimes Q \to L_0 \rtimes P, (x,B) \mapsto (\alpha(x),\beta(B))$$
is an isomorphism. Thus $G_Q = L_0 \rtimes Q \cong L_0 \rtimes P = G_P$.
Next I show that $G_Q$ is isomorphic to a symmorphic space group. By Maschke's theorem
$Q$ is conjugate in $GL_n(\mathbb{R})$ to a subgroup $P$ of $O_n$. So there is
$A \in GL_n(\mathbb{R})$ such that $P = AQA^{-1}$. Let $L := AL_0 \le \mathbb{R}^n$. Then the isomorphisms $\alpha: L_0 \to L$, $\beta: Q \to P$ (defined by the same formulars as above) yield $G_Q \cong L \rtimes P \cong \langle t_a \mid a \in L \rangle \rtimes \langle r_B \mid B \in P \rangle$ and the latter is the sought-after symmorphic space group.
1. Step: $f$ is injective
Let be $f([Q_1]) = f([Q_2])$ and write $G_i := G_{Q_i}$. Then there is an
isomorphism $\phi: G_1 \to G_2$ of groups and we wan't to show that
$Q_1, Q_2$ are conjugate in $GL_n(\mathbb{Z})$.
From the lemma in Step 3 it follows that $L_0$ is the unique maximal free abelian rank $n$ subgroup of each $G_Q$. Since maximality is preserved by group isomorphisms,
$\phi(L_0) \cong L_0$ is a maximal free abelian subgroup of rank $n$ of $G_2$
and by uniqueness, $\phi(L_0) = L_0$. Now by a general principal in group
theory, $\phi$ induces an isomorphism $\bar{\phi}: Q_1 \to Q_2$ such that
there is a commutative diagramm:
$$ 1 \to L_0 \to G_1 \overset{\kappa_1}{\to} Q_1 \to 1 $$
$$ \hspace{5pt} \phi \downarrow \hspace{15pt} \phi \downarrow \hspace{17pt} \downarrow\bar{\phi} \hspace{15pt}$$
$$ 1 \to L_0 \to G_2 \underset{\kappa_2}{\to} Q_2 \to 1 $$
$\bar{\phi}$ can be described as follows: to $B \in Q_1$ choose
$X \in G_1$ such that $\kappa_1(X) = B$. Then
$\bar{\phi}(B) = (\kappa_2 \circ \phi)(X)$. (In the current situation we can always take $X=(0,B)$ ).
Since $\phi \in Aut(L_0) = Aut(\mathbb{Z}^n)$, there is
$A \in GL_n(\mathbb{Z})$ such that $\phi(a,I) = (Aa,I)$.
For $B \in Q_1$ write $\phi(0,B) = (b,C)$ with an $C \in Q_2$. Let $e_i$ be a standard vector. From
$$(CAe_i,I)= (b,C) \cdot (Ae_i,I) \cdot (b,C)^{-1} = \phi((0,B) \cdot (e_i,I) \cdot (0,B)^{-1})$$
$$ = \phi(Be_i,I)= (ABe_i,I)\hspace{80pt}$$
one finds $CA=AB$, i.e. $C=ABA^{-1}$. This yields
$$\bar{\phi}(B) = (\kappa_2\circ \phi)(0,B)= \kappa_2(b,ABA^{-1}) = ABA^{-1},$$ hence $\bar{\phi}$
is conjugation with $A$ and $Q_2= \bar{\phi}(Q_1) = AQ_1A^{-1}$ is
conjugated to $Q_1$ in $GL_n(\mathbb{Z})$.
Step 2: $f$ is surjective
Let $G$ be a symmorphic space group. We have to show that there exists
a finite subgroup $Q \le GL_n(\mathbb{Z})$ such that $G \cong G_Q$.
Let $T$ be the translational subgroup of $G$ with lattice
$L = \oplus_{i=1}^n \mathbb{Z}a_i$. Set $A := (a_1,...,a_n) \in GL_n(\mathbb{R})$.
Then $\alpha: L_0 \to L, x \to Ax$ is an isomorphism that induces
a further isomorphism
$$\phi: GL_n(\mathbb{Z}) = Aut(L_0) \to Aut(L), B \mapsto ABA^{-1}.$$
Now let be $P := \mathcal{Q}(L) \le Aut(L) \cap O_n$ (notation in OP in 1.).
In particular, as a discrete subgroup of a compact group, $P$ is finite. Thus $Q := \phi^{-1}(P)$ is a finite subgroup of $GL_n(\mathbb{Z})$ and $\beta: Q \to P, B \mapsto \phi(B) = ABA^{-1}$ is an isomorphism.
Since $G$ is symmorphic, $G = L \rtimes P$ where $P$ acts on $L$ via matrix
multiplication. By definition, $G_Q = L_0 \rtimes Q$ where $Q$ also
operates via matrix multiplication. Now $G \cong G_Q$ follows exactly
as in Step 0.
Step 3: This result was used in Step 1.
Lemma: The translational subgroup is the unique maximal free abelian rank $n$ subgroup of a space group of dimension $n$.
Let $T$ be the translational subgroup of the space group $G$ (dimension $n$) and let $F \le G$ be a free abelian subgroup of rank $n$. From finiteness of $G/T$ follows that $TF/T \cong F/F \cap T$ is finite. Thus $F \cap T$ is free abelian of rank $n$, hence $F \cap T$ has finite index in $T$. Let $t_{a_i} = (a_i,I)$ $(i=1,...,n)$ be basis vectors for $T$. Then we can choose $k \in \mathbb{Z}, k \neq 0$ such that $(t_{a_i})^k = (ka_i,I) \in F$.
Let $(a,A) \in F$. Since $F$ is abelian, $(a,A)$ and $(ka_i,I)$ commute, yielding $(a+Aka_i,A)=(ka_i+a,A)$ and in particular $Aa_i = a_i$. Thus $A=I$, i.e. $(a,A)$ is a translation, which means $(a,A) \in T$. Hence $F \le T$.
Next I show that $T$ is maximal free abelian of rank $n$: Let $F$ be as above and assume $T \le F$. Since $F \le T$ as shown before, $T=F$ follows. Thus $T$ is maximal.
Finally I show that $T$ is the only maximal free abelian subgroup of rank $n$. Let $F$ be another one. Since again $F \le T$, maximality of $F$ implies $F=T.\quad\quad$ q.e.d.
Best Answer
Such triples, involving two objects -or even categories- and some kind of an "action" of one of them on the other (respecting some or all of its structure maps), are quite general and are met -as has already been indicated in the other answers and the comments as well- in various different branches of mathematics and mathematical physics. A fair coverage might deserve a suitably sized expository article.
I will try to sketch some of these ideas, and the corresponding objects, but it is important to note that there are lots of others as well (just to mention a few: $C^*$-algebras, graded algebras, etc, provide somewhat similar examples in different directions).
$\blacksquare$ Case $I$: The smash product $A\sharp H$, between a hopf algebra $H$ and a $H$-module algebra $A$
One notion which is similar and in some sense generalises the first construction discussed in the OP, is the smash product algebra, between a Hopf algebra $H$ and an algebra $A$, which also a $H$-module algebra:
Now what might be particularly interesting, is that the above smash product algebra, generalizes some similar but well known constructions. So let us study closer some specific cases of smash product algebras:
Let $Α$ be a (left) $kG$-module algebra. Using the definition above, we get that for all $g \in G$ we have: $$ g\cdot (ab) = (g \cdot a)(g \cdot b) $$ i.e.: the elements of $G$ act as automorphisms of $Α$. We thus get a group homomorphism: $$ \sigma: G \longrightarrow Aut_{k}A $$ and conversely: given any automorphism $\sigma: G \rightarrow Aut_{k}A$, $Α$ becomes a (left) $kG$-module algebra. By the definition of the multiplication in $A \sharp kG$ : $$ (a \sharp g)(b \sharp h) = a(g \cdot b) \sharp gh = a \sigma(g)(b) \sharp gh $$ which proves that, in this case the smash product algebra is isomorphic to the skew group algebra: $$A \sharp kG \cong A \star G$$ as $k$-algebras.
The next example, may be seen as a special case of the above:
Let $Κ$ and $Η$ two groups and let us assume that the elements of $K$ act as automorphisms of $H$. (We thus have a group homomorphism: $\phi: Κ \rightarrow Aut(H)$). Thus, the elements of $Κ$ act as automorphisms of the group algebra $kH$, which becomes a $kK$-module algebra. From the definition of the multiplication in $kH \sharp kK$ we get that for all $h, h' \in H$ και $g, g' \in K$: $$ (h \sharp g)(h' \sharp g') = h(g \cdot h') \sharp gg' = h \phi(g)(h') \sharp gg' $$ i.e., in this case: $$ kH \star K \cong kH \sharp kK \cong k(H \rtimes_{\phi} K) $$ where $H \rtimes_{\phi} K$ stands for the semidirect product of the groups $Η$, $Κ$ (which is a group itself) and $k(H \rtimes_{\phi} K)$ is its group algebra.
A remarkable consequence, is that, if the action $\phi$ is trivial, in the sense that, for all $g \in K$ and for all $h \in H$: $$ g \cdot h = \phi(g)(h) = h $$ then the semidirect product is identified with the direct product (as groups) and the smash product is identified with the tensor product (as algebras). In that case, the former relation reproduces the well known (from group representation theory): $$ kH \otimes kK \cong k(H \times K) \equiv k(H \oplus K) $$
On the other hand, the smash product algebra described above, plays a much more important role in the structure theory of hopf algebras than simply generalizing or mimicking other known similar constructions: For example, it is well known that any cocommutative hopf algebra $H$ over an algebraically closed field of characteristic zero is isomorphic to the smash product hopf algebra between the group hopf algebra of its grouplikes $G(H)$ and the UEA of the Lie algebra of its primitives $U(P(H))$: $$ H\cong U(P(H))\sharp kG(H) $$ This is commonly refered to as the Cartier-Kostant-Milnor-Moore theorem.
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$\blacksquare$ Case $II$: The crossed product $A\sharp_\sigma kG$, between an algebra $A$ and a finite group $G$
Another similar construction of crossed products, not -in principle- related with the Hopf algebra case and its module algebra described in the former case, is the following:
Let $A$ be a finite dimensional algebra over an algebraically closed field and $G$ a finite group.
Let also, the following additional data:
$\bullet$ a convolution-invertible map $\sigma:G\times G\rightarrow A$ (Note that: $\sigma\in Hom_k(kG\otimes kG,A)$)
$\bullet$ a $k$ linear map $kG\otimes A\rightarrow A$, denoted $g\otimes a\mapsto g\cdot a$
satisfying: $$ g\cdot(h\cdot a)= \sigma(g,h)(gh\cdot a)\sigma^{-1}(g,h) \rightsquigarrow twisted \ module \ condition \\ \big(g\cdot\sigma(h,l)\big)\sigma(g,hl)=\sigma(g,h)\sigma(gh,l) \rightsquigarrow cocycle \ condition \\ g\cdot(ab)=(g\cdot a)(g\cdot b) \rightsquigarrow kG \ measures \ A \\ \sigma(1,g)=\sigma(g,1)=1, \ \ g\cdot1=1, \ \ 1\cdot a=a $$ for all $g,h,l\in G$ and for all $a,b\in A$. Thus, $\sigma$ is a $2$-cocycle.
It is a fully $G$-graded algebra, with $g$-component: $(A\sharp_\sigma kG)_g=A\sharp_\sigma kg$ for all $g\in G$ and with the algebra $A$ embedding in it as the identity component. You can see some discussion about it in the article Irreducible representations of crossed products, J. of Pure and applied algebra, v.129, 3, p. 315-326, 1998, by S. Montgomery and S.J. Witherspoon.
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$\blacksquare$ Generalization of Cases $I$ and $II$: The crossed product $A\sharp_\sigma H$, between an algebra $A$ and a hopf algebra $H$, which "measures" $A$
We will say that a hopf algebra $H$ measures the algebra $A$, if there is a $k$-linear map $H\otimes A\to A$, given by $h\otimes a \mapsto h\cdot a$ and such that $h\cdot 1=\epsilon(h)1$ and $h\cdot (ab)=\Sigma(h_{(1)}\cdot a)(h_{(2)}\cdot b)$ for all $h\in H$, $a,b\in A$.
Note that now, unlike case $I$, we do not demand $A$ to be an $H$-module. Based on the above definition, the next lemma can be shown in a relatively straightforward manner:
In case $\sigma$ is trivial, in the sense that $\sigma(h,k)=\epsilon(h)\epsilon(k)1$ for all $h,k\in H$, then the above conditions directly imply that $A$ is an $H$-module algebra and that the crossed product becomes the smash product $$ A\sharp_\sigma H=A\sharp H $$ of case $I$, discussed above.
In case $H=kG$, where $G$ is a finite group, then the crossed product becomes the group crossed product $$ A\sharp_\sigma H=A\sharp_\sigma kG $$ of case $II$, discussed above.
(More details on the properties of the structure and representations of the crossed product algebra $A\sharp_\sigma H$ can be found in ch. 7 of S. Montgomery's book "Hopf algebras and their actions on rings").