Is there a name of semidirect product of a group with its automorphism group? Consider the construction $G \rtimes \text{Aut}(G)$. Here $
G$ is a group, $\text{Aut}(G)$ is the automorphism group and the semidirect product is over the most obvious action.
1) Is there any name for such a general construction? To me, it seems like the most straight-forward example of a semi-direct product.
2) Are there any surveys over such constructions or any big theorems about the structure of such groups? 
3) I'm specifically interested in finding torsional elements when $G = F_2$, the free group of two generators. Is there any result about this special scenario?
Edit: More thoughts about question 3 are below.
As pointed in the comments, if you have any torsional automorphism $\phi$ of the free group (there are good classification theorems for such automorphisms), then a torsional element of the semidirect product will be of the form $(g,\phi)$ such that $g\phi(g)\phi^2(g)... \phi^{k-1}(g)=1$, $k$ being the order of $\phi$.
You can check that any element of the form $(\alpha^{-1} \phi(\alpha),\phi)$ works where $\phi$ is a torsional automorphism and $\alpha \in F_2$. Are these all such elements? Is there a general form of such elements?
 A: As a first remark, note that if $\tilde{H}\leq G\rtimes \operatorname{Aut}(G)$ is a finite subgroup and $G$ is torsion-free, then the projection $p: G\rtimes \operatorname{Aut}(G)\to \operatorname{Aut}(G)$ maps $\tilde{H}$ isomorphically to some finite subgroup $H=p(\tilde{H})\leq \operatorname{Aut}(G)$. 
Now for each finite subgroup $H\leq \operatorname{Aut}(G)$, you can ask yourself what are the subgroups $\tilde{H}\leq G\rtimes \operatorname{Aut}(G)$ projecting to $H$? These correspond to splittings of the semi-direct product $G\rtimes H$, which correspond to $1$-cocycles (or crossed homomorphisms) $f:H\to G$, that is, functions satisfying 
$$
f(ab) = f(a){}^af(b),\quad a,b\in H.
$$ 
Here ${}^ag$ denotes the action of $a$ on $g$. These $1$-cocycles represent elements of the first non-abelian cohomology set $H^1(H;G)$, the trivial element of which is represented by any cocycle of the form $f(a) = (g^{-1}){}^ag$ for some $g\in G$.
So in some sense the answer to your follow-up question lies in the non-abelian cohomolgy of finite subgroups of $\operatorname{Aut}(G)$.
