Torsion-free Aut(G), but torsion in Out(G) Context
I am trying to understand how group extensions with non-abelian kernels can be classified. Chapter IV.6. of Ken Brown's Cohomology of groups gives a sketch of this using crossed-modules and cohomology.
I says that any extension $1 \to N \to E \to G\to 1$ gives rise to a canonical homomorphism $\psi:G\to \text{Out}(N)$. However, not every such homomorphism comes from a group extension.
I am trying to find an example where indeed $\psi$ doesn't come from a group extension. One case where this won't be the case is when $\psi$ has a lift to $\varphi : G \to \text{Aut}(N)$. To make sure this doesn't happen...
The Question
I am looking for a group $G$ such that:

*

*$\text{Aut}(G)$ is torsion-free, and

*$\text{Out}(G)$ has torsion.

Do any examples of such come to your mind?
 A: Theorem. For every finite group $F$ there exists a torsion-free finitely-presented group $G$ (which is actually hyperbolic, if you care about such things) such that:
(a) $Aut(G)$ is torsion-free.
(b) $Out(G)\cong F$.
Proof. For the purpose of this answer, an $n$-dimensional hyperbolic manifold is a connected compact manifold (without boundary) $M$ which is homeomorphic to the quotient of the $n$-dimensional hyperbolic space $H^n$ by a torsion-free discrete group of isometries $\Gamma$. In this case, $\Gamma\cong \pi_1(M)$. This homeomorphism defines a hyperbolic metric on $M$, which is a Riemannian metric of constant sectional curvature $-1$.
The key property to know is that, whenever $n\ge 3$, the isometry group of $M$ is finite and is isomorphic to $Out(\Gamma)$ (this the a consequence of the Mostow Rigidity Theorem).
It was proven by Kojima for $n=3$ and by Belolipetsky and Lubotzky for $n\ge 4$ that for every finite group $F$, there exists a hyperbolic $n$-dimensional manifold $M$ whose isometry group is isomorphic to $F$:
Kojima, Sadayoshi, Isometry transformations of hyperbolic 3-manifolds, Topology Appl. 29, No. 3, 297-307 (1988). ZBL0654.57006.
Belolipetsky, Mikhail; Lubotzky, Alexander, Finite groups and hyperbolic manifolds, Invent. Math. 162, No. 3, 459-472 (2005). ZBL1113.57007.
A closer look at their proofs shows a bit more: The isometry group $Isom(M)$ of $M$ acts freely on $M$, i.e. only the identity element of $Isom(M)$ has fixed points in $M$.
Now, given a finite group $F$, let $M$ be a hyperbolic $n$-dimensional manifold whose existence is guaranteed by the results of Kojima, Belolipetsky and Lubotzky (of course, Kojima's theorem suffices).
I claim that $G\cong \pi_1(M)$ satisfies the properties stated in the theorem. The group $G$ is isomorphic to a discrete torsion-free subgroup $\Gamma$ of isometries of $H^n$ such that $M\cong H^n/\Gamma$.
Indeed, $F\cong Isom(M)\cong Out(F)$, proving (b). In order to prove (a) one observes (this is again a consequence of the Mostow Rigidity Theorem) that the group $Aut(G)$ is isomorphic to the normalizer $N(\Gamma)$ of $\Gamma$  in the isometry group of the hyperbolic space $H^n$, which, in turn, equals the group of all lifts of elements of $Isom(M)$ to $H^n$: Given $\phi\in Isom(M)$, an isometry $\tilde\phi$ of $H^n$ is called a lift of $\phi$ if $\pi\circ \tilde\phi= \phi\circ \pi$, where $\pi: H^n\to M$ is the quotient map. I claim that this normalizer $N(\Gamma)$ is torsion-free. Indeed, if $\tilde\phi\in N(\Gamma)$ is a finite-order element, it has at least one fixed point $\tilde{x}\in H^n$. Then $\pi(\tilde{x})\in M$ is fixed by $\phi$. However, as noted above, $Isom(M)$ acts on $M$ freely. Hence, $\tilde\phi$ is a lift of the identity map of $M$. Thus, $\tilde\phi\in \Gamma$. But the group $\Gamma$ is torsion-free, hence, $\tilde\phi=1$. Thus, $N(\Gamma)\cong Aut(\Gamma)\cong Aut(G)$ is torsion-free. qed
If you never worked with hyperbolic manifolds before, all this is probably hard to digest. But these are the examples that I know.
Edit. Here is the algebraic form of the Mostow Rigidity Theorem (MRT) used in my answer:
For each $n\ge 3$ the following holds. Suppose that $\Gamma_1, \Gamma_2< Isom(H^n)$ are two discrete subgroups with quotients $H^n/\Gamma_1, H^n/\Gamma_2$ of finite volume. Then for every isomorphism $\rho: \Gamma_1\to \Gamma_2$ there exists (a unique) isometry $\psi=\psi_{\rho}\in Isom(H^n)$ which induces $\rho$ in the sense that
$$
\rho(\gamma)= \psi \circ \gamma \circ \psi^{-1}, \forall \gamma\in \Gamma_1.
$$
In the case of interest, $\Gamma_1=\Gamma_2=\Gamma$ and $\rho\in Aut(\Gamma)$. Then the MRT states that there exists (a unique) $\psi\in N(\Gamma)$ such that $\rho(\gamma)=  \psi \circ \gamma \circ \psi^{-1}$ for all $\gamma\in \Gamma$.
Thus, one obtains a natural isomorphism $Aut(\Gamma)\to N(\Gamma)$, $\rho\mapsto \psi_\rho$. Taking the quotient by inner automorphisms of $\Gamma$, one obtains the isomorphism
$$
Out(\Gamma)\to N(\Gamma)/\Gamma\cong Isom(H^n/\Gamma).
$$
A: This doesn't answer your question (I can't even think of a group with a nontrivial torsion-free automorphism group), but I can give you example of a map $\psi:G \to {\rm Out}(N)$ that does not give rise to and extension $E$.
Let $N = {\rm SL}(2,9) = 2.A_6$. Then ${\rm Aut}(N)$ is isomorphic to the full automorphism group of $A_6$, which is ${\rm P \Gamma L}(2,9) = A_6.2^2$, and ${\rm Out}(N) \cong C_2^2$ (a Klein $4$-group).
Let $G$ have order $2$. Then there are three nontrivial homomorphisms $G \to {\rm Out}(N)$.
Two of these (the ones arising from $S_6$ and ${\rm PGL}(2,9)$) give rise to extensions $E = 2.A_6.2$, but the third one (corresponding to ${\rm P \Sigma L}(2,9) = M_{10}$) does not - there is no group with shape $2.M_{10}$.
