Groups which maintain all their subgroups’ automorphisms as inner automorphisms Are there any groups, finite or infinite, other than the first three symmetric groups which maintain all their subgroups’ automorphisms as inner automorphisms (every automorphism of every subgroup extends to an inner automorphism of the whole group)?
EDIT: I just had an idea. Does the Rado graph’s automorphism group work?
 A: Yes. By this paper of Minasyan one can construct a finitely generated group $G$ with all proper subgroups cyclic of prime order $p\gg 1$, two conjugacy classes and trivial $Out(G)$. This group obviously satisfies the conditions of OP.
A: The answer to your question is no for finite groups. That is there are no other finite examples that satisfy your condition.
A finite $p$-group $P$ with $\lvert P\rvert>p$ has an outer automorphism of $p$-power order. That is a well-known result of Gaschütz for nonabelian $P$, and is easily checked for abelian $P$.
So all Sylow $p$-subgroups of a finite group $G$ satisfying your condition must be cyclic of order $p$. Then, by repeated application of Burnside's Transfer Theorem to the primes dividing $\lvert G\rvert$ in increasing order, we find that $G$ has a normal Sylow $p$-subgroup $P$, where $p$ is the largest prime dividing $\lvert G\rvert$.
Your condition implies that $N_G(P)/C_G(P)=G/C_G(P)$ is isomorphic to $\operatorname{Aut}(P)$, which is cyclic of order $p-1$.
Now, if $p>3$, then (since $p-1$ is square-free), there is some prime $q$ with $2<q<p$ and $q \mid p-1$, and there exists $N \lhd G$ with $C_G(P) < N$ and $\lvert G/N\rvert=q$, so $G=NQ$ with $Q \in \operatorname{Syl}_q(G)$. But now $N_G(Q) = N_N(Q)Q = C_G(Q)$, so the automorphisms of $Q$ cannot be induced by elements of $N_G(Q)$, contrary to assumption.
Hence $p \le 3$, and $G=\operatorname{Sym}(n)$ with $n=1$, $2$ or $3$.
