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)?

5$\begingroup$ I take it that by "maintain all their subgroups' automorphisms as inner automorphisms" you mean for any subgroup $H\subseteq G$, and any automorphism $\varphi$ of $H$, we have an automorphism $\widetilde{\varphi}$ of $G$ such that $\widetilde{\varphi}$ restricted to $H$ is $\varphi$, and such that $\widetilde{\varphi}$ is inner. Is that right? $\endgroup$ – Sam Hopkins Jun 10 at 17:37

2$\begingroup$ Please say what you mean by "maintain all their subgroups' automorphisms as inner automorphisms". Why should other people have to try and guess what you might mean? $\endgroup$ – Derek Holt Jun 10 at 18:04

$\begingroup$ Whatever the question means, cyclic groups of prime order are surely among the groups sought. $\endgroup$ – LSpice Jun 10 at 18:43

2$\begingroup$ @LSpice probably not, because they are not complete groups for odd primes. $\endgroup$ – Derek Holt Jun 10 at 18:49

3$\begingroup$ Equivalently. $N_G(H)$ maps onto ${\rm Aut}(H)$ for all subgroups $H \le G$. $\endgroup$ – Derek Holt Jun 10 at 20:50
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 wellknown 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 $p1$.
Now, if $p>3$, then (since $p1$ is squarefree), there is some prime $q$ with $2<q<p$ and $q \mid p1$, 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 $3$.
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.

1$\begingroup$ I don't understand this example. How can the automorphisms of the subgroups of order $p$ be induced by inner automorphisms of $G$? $\endgroup$ – Derek Holt Jun 11 at 7:26

1$\begingroup$ @DerekHolt the nontrivial subroups $H$ are cyclic, and any two nontrivial elements of $H$ are conjugate in $G$. $\endgroup$ – Mikael de la Salle Jun 11 at 7:30

4$\begingroup$ @DerekHolt Ah! Theorem 1.1 of Minasyan's paper implies that a Tarski monster for prime $p$ can be embedded in a $2$generated group where every element of finite order has order $p$ and all elements of order $p$ are conjugate. But it doesn't claim that this group has no elements of infinite order. $\endgroup$ – Jeremy Rickard Jun 11 at 8:58

2$\begingroup$ @JeremyRickard Yes, and all elements of infinite order are conjugate. But how do we deduce from results in the paper that this example has trivial outer automorphism group? $\endgroup$ – Derek Holt Jun 11 at 9:18

4$\begingroup$ @MikaeldelaSalle I'm afraid my work can't be used to answer this question. I do not think that there is a group with the property that every proper subgroup is cyclic and any two nontrivial elements are conjugate (other than $\mathbb{Z}/2\mathbb{Z}$). Indeed, if $x \neq 1$ and $yxy^{1}=x^{1}$ then $y$ will normalize $\langle x \rangle$, hence the subgroup $\langle x,y\rangle$ will be metabelian but noncyclic (so, necessarily proper). $\endgroup$ – Ashot Minasyan Jun 11 at 15:12