Finite-extensible automorphisms of finite groups are inner. The proof is rather non-trivial. For the classes of finite nilpotent and finite soluble groups, this is proved in a paper of Pettet, "On inner automorphisms of finite groups". It can be extended to the class of all finite groups using a theorem of Hartley and Robinson, "On finite complete groups". For how to assemble the proof, see the wiki page https://groupprops.subwiki.org/wiki/Finite-extensible_implies_inner
Edit: What the above shows is that given $G$, there is a single complete group $K$ and an inclusion $G\to K$ with the property that the only automorphisms of $G$ that extend to automorphisms of $K$ are inner ones. The group $K$ in question can be taken to be a semidirect product $P\rtimes G$ where $P$ is a finite $p$-group and $p$ is any given prime that does not divide $|G|$.
This does not quite prove what Noah is asking for here, because what he calls an extendable automorphism is really an extension of the automorphism. So there remains the question of whether the identity automorphism of $G$ extends in a non-trivial way.
If $G$ is trivial then we can prove that this cannot happen as follows. The map $\beta$ assigns to every finite group an automorphism in a way that is compatible with group homomorphisms. So by the above theorem it is inner, and takes every subgroup of a group to itself. Moreover, an inner automorphism of a cyclic group is the identity, so $\beta$ is the identity. However, it does not seem so easy to prove this when $G$ is not trivial, so this is an incomplete answer at this point.
Note that by contrast, in the category of finite abelian groups, the trivial automorphism of the trivial group has $2^{\aleph_0}$ extensions to all finite abelian groups. Namely, for each prime, you have two choices: negate or don't negate.