TITLE: Groups with given automorphism groups
QUESTION [21 upvotes]: It is an easy exercise to show that all finite groups with at least three elements have at least one non-trivial automorphism; in other words, there are - up to isomorphism - only finitely many finite groups $G$ such that $Aut(G)=1$ (to be exact, just two: $1$ and $C_2$).
Is an analogous statement true for all finite groups? I.e., given a finite group $A$, are there - again up to isomorphism - only finitely many groups $G$ with $Aut(G)\cong A$?
If yes, is there an upper bound on the number of such groups $G$ depending on a property of $A$ (e.g. its order)?
And if not, which groups arise as counterexamples?
And finally, what does the situation look like for infinite groups $G$ with a given finite automorphism group? And what if infinite automorphism groups $A$ are considered?

REPLY [3 votes]: A more accessible account on the theorem of Ledermann-Neumann mentioned in the accepted answer can be found here: Math. Monthly or arxiv
EDIT: The argument runs as follows. Suppose $G$ is a finite group with exactly $n$ automorphisms.

The order of the derived group $G'$ is bounded in terms of $n$. Since the size of the inner automorphism group $|G/Z(G)|$ is bounded by $n$, this follows from a theorem of Schur. While Schur's proof is highly non-trivial, there is an elementary argument by Rosenlicht.
Every prime divisor of $|G|$ is bounded by $n+1$. This requires a special case of the Schur-Zassenhaus theorem for central Sylow subgroups.
The exponent of $G$ is bounded in terms of $n$. This uses an elementary argument by Nagrebeckiı̆.
The size of the center $Z(G)$ is bounded in terms of $n$. Since the exponent is bounded, it suffices to bound the minimal number of generators of $Z(G)$. Let $g_1,\ldots,g_m$ representatives for the cosets of $G$ mod $Z(G)G'$ and consider $U:=\langle g_1,\ldots,g_m\rangle G'$. Note that $m$ is bounded in terms of $n$. Using another lemma, one finds a decomposition $Z(G)=C\times D$ such that $U\cap Z(G)\le C$ and $|C|$ is bounded. It remains to bound $|D|$. The author shows that $G=UC\times D$. Now we may assume that $G=D$ is an abelian $p$-group. It is easy to see that the number of automorphisms of $G$ grows with its rank.