Is there any $k > 2$ such that if $\mathbb Z_k$ is subgroup of $H$, then $\operatorname{Aut}(\mathbb Z_k) = \mathbb Z_{\phi(k)}$ is subgroup of $\operatorname{Aut}(H)$? What if we require $H$ to be finite?
The original question was if it is true for all $k$, but it is not. There is group $H$ s.t. $|\operatorname{Aut}(H)|$ is odd (and greater than $1$). This group has cyclic subgroup of order greater than $2$, and such a group has an even number of automorphisms, and thus isn't subgroup of $\operatorname{Aut}(H)$.
If we require $H$ to be finite abelian, then there is no such $k$, as any automorphism of $\mathbb Z_k$ can be extended to automorphism of $H$.
This doesn't work at all for any $k$ for arbitrary $H$, as Schupp ("A Characterization of Inner Automorphisms") proved that if automorphism of subgroup extends to automorphism of any larger group then the original automorphism is inner. However, that of course doesn't mean that automorphisms of the larger group don't contain automorphisms of our group as a subgroup in some way other than continuation.