No, a characteristic subgroup of an automorphism group of a group need not be isomorphic to the automorphism group of any group.
Here are two very well known infinite families of examples:
The cyclic group of order p (p an odd prime) is not the automorphism group of any group (such a group is abelian since its inner automorphism group is cyclic, but an abelian group either has inversion as an automorphism of order 2, or is an elementary abelian 2-group, and an elementary abelian 2-group either has trivial automorphism group or has a coordinate swap as an order 2 automorphism). However, the cyclic group of order p is a characteristic Sylow p-subgroup of AGL(1,p), which is the automorphism group of the dihedral group of order 2p and of itself. AGL(1,p) is the normalizer of a Sylow p-subgroup of the symmetric group on p-points.
The alternating group of degree n (n ≥ 9) is a characteristic, index 2 subgroup of its automorphism group, but is not itself the automorphism group of any group by Robinson (1982, MR678545).
At least as far as I understand it, automorphism groups of groups tend to be big and "full", and so it should not be surprising that many of their subgroups are not themselves automorphism groups of groups since they are "missing" something. For instance, a simple group cannot be an automorphism group unless it is complete; M12 is incomplete. Odd order cyclic groups do not work, since one is missing inversion (and indeed, the rest of AGL1).
Ok thanks to the reference given in the comment, I can give the answer to those questions.
First a corollary from a result in " V. T. NAGREBECMI. On the periodic part of a group with finitely many automorphisms.
Doll. Akad. Nauk. SSSR 205 (1972). 519-521: Sol,ier Math. DoXI. 13 (1972). 953-956."
If $G$ is an infinite abelian group with $|Aut(G)|<\infty$ then there exists a subgroup $F$ without torsion of $G$ such that $G=Tor(G)\oplus F$ and $Tor(G)$ is a finite group.
Hence we see that to answer question 1 it suffices to deal with infinite abelian groups without torsion, but with the following citation taken in J. T. HALLETT AND K. T. HIRSCH. Torsion-free groups having finite automorphism
groups, I. J. Algebra 2 (1965). 287-298., we see that it is hard :
"we do not concern ourselves with the apparently hopeless task of finding all the torsion free abelian groups whose automorphism group is a given finite group"
Following their comment, we can say that 1. is "hopeless".
However question 3 admits an answer. The answer is yes if $A$ is the center of $A$. The proof goes as follow, any $\phi\in Aut(G)$ induces an automorphism of $\varphi\in Aut(Z(G))$ and an automorphism $\psi\in Aut(Q)$ if $\iota$ is the inclusion of $Z(G)$ in $G$ and $\pi$ the projection of $G$ onto $F$ the automorphisms verify :
$$\iota\circ\varphi=\phi\circ\iota\text{ and } \pi\circ \phi=\psi\circ \pi $$
On the whole this gives a group morphism :
$$\mathcal{A}: Aut(G)\rightarrow Aut(Z(G))\times Aut(F) $$
Furthermore its kernel $Ker(\mathcal{A})$ is easily seen to be isomorphic to $Hom(F,Z(G))=Hom(F^{ab},Z(G))=Hom(F^{ab},Tor(Z(G)))$.
Assume that $Aut(Z(G))$ is finite then by the result above $Tor(Z(G))$ is finite hence $Ker(\mathcal{A})$ is finite and because $Im(\mathcal{A})$ is in a finite subgroup, it must be finite as well, on the whole $Aut(G)$ is finite.
Now for the question 2, I did not find a counter-example but it should be false.
Best Answer
Ledermann and B.H.Neumann ("On the Order of the Automorphism Group of a Finite Group. I", Proc. Royal Soc. A, 1956) have shown the following:
An immediate consequence is that up to isomorphism, there are only finitely many finite groups $G$ with $|\operatorname{Aut}(G)| \leq n$. Hence for any finite group $X$, up to isomorphism there are only finitely many finite groups $G$ with $\operatorname{Aut}(G) \cong X$.
Among infinite groups this is no longer true, and indeed there are infinitely many groups $G$ with $\operatorname{Aut}(G) \cong \mathbb{Z} / 2 \mathbb{Z}$.
Then there is of course the question of determining all finite groups $G$ with given automorphism group $\operatorname{Aut}(G) \cong X$. For this, see for example
This paper gives a solution to the problem in some cases, and determines for example all $G$ with $\operatorname{Aut}(G) \cong S_n$. There is also a different proof of the fact that there are only finitely many groups with a given automorphism group (Theorem 3.1 there).