I was reading [This question on MO](https://mathoverflow.net/questions/36175/is-there-any-criteria-for-whether-the-automorphism-group-of-g-is-homomorphic-to) and suddenly faced another question.


Is there any necessary and sufficient condition for $G$ such that there exist a group $G'$ that $G \cong Aut(G')$ ?
What can we say about subgroups of $G$ when we know the group $G'$ with that condition exist.



Is there any criterion for whether the group G is isomorphic to the automorphism group of another group?
Let me expand on my comment. There are two papers that you should look at:



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> Iyer, Hariharan K. *On solving the equation $Aut(X)=G$*. 
>  Rocky Mountain J. Math. 9 (1979), no. 4, 653–670. 
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The scope of this paper is when your group $G'$ is **finite**. The MathSciNet review gives a perfect precis:



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> The author considers the finite groups having a given automorphism group $H$. First he shows (Theorem 3.1) that for every finite group H there are only finitely many finite groups $G$ having $H$ as automorphism group, and he surveys the possible groups $G$ in the cases when $H$ is one of the following types of group: alternating, symmetric, dihedral, dicyclic, quasidihedral (Theorems 4.2, 4.4, 6.3 and 6.6).
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The second paper you should consult does not restrict $G'$ to being finite:



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> Robinson, Derek J. S. *Groups with prescribed automorphism group.*
>  Proc. Edinburgh Math. Soc. (2) 25 (1982), no. 3, 217–227. 
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Again the MSN review tells the story:



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> The author treats the question of whether a given finite group is the full automorphism group of at least one group. The main theorem gives a description of all groups $G$ such that $Aut(G)$ is finite and $Aut\_c(G)$, the subgroup of central automorphisms, is semisimple. Using this, he shows that for a finite simple (nonabelian) group $S$ there is a group $G$ such that $Aut(G)≅S$ if and only if one of the following is true: (a) $S=PSL(r,2)$, $r>2$, and $G$ is elementary abelian, (b) $S$ is complete, and $G\cong \hat{S}/K$ or $(\hat{S}/K)\times Z\_2$, where $K$ is any subgroup of $M(S)$, the Schur multiplier of $S$, and $|M(S):K|$ is odd in the second case, (c) $S=Sz(8)$ and $G\cong\hat{S}/K$, $K<M(S)$, $|K|=2$..... For a finite group $S$ there are only finitely many finite groups G with $Aut(G)\cong S$. This is false if $G$ is infinite. For $S\cong S\_4$ the author constructs uncountably many infinite nonabelian groups G with $Aut(G)\cong S\_4$.
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Note that the result detailing which simple groups are automorphism groups of an arbitrary group suggests that a general necessary and sufficient condition seems unlikely to exist.


