Solving $|Aut(H)| = |Aut(G)|$

Couple of days ago, i actually asked this question : Existence of a normal subgroup with $|\operatorname{Aut}{(H)}|>|\operatorname{Aut}{(G)}|$.

I was thinking about the converse of this statement. Suppose $G$ is a finite group, with subgroups $H$ and satisfies $$|Aut(G)|=|Aut(H)|,$$ then can anything be said about, $H$ or $G$ in terms of their structure. Will $H$ be normal or ....? Or one can even ask, that find all finite groups $G$ such that $|Aut(H)|=|Aut(G)|$ for every subgroup $H$ of $G$! I am not interested that much in this question as there seems one can have lot of Groups of this type, but i am more curious to know the behaviour of $H$.

Also, anyone interested can very well read this article in MO: https://mathoverflow.net/questions/1075/order-of-an-automorphism-of-a-finite-group

In response to:

...find all finite groups $G$ such that $|\mathrm{Aut}(H)|=|\mathrm{Aut}(G)|$ for every subgroup $H$ of $G$

the only groups (finite or infinite) for which this is true is the trivial group and the group of order $2$.

Proof: All groups have the trivial group $\{1\}$ as a subgroup, which has a trivial automorphism group, i.e., $|\mathrm{Aut}(\{1\})|=1$. By the assumptions on $G$, we have $|\mathrm{Aut}(\{1\})|=|\mathrm{Aut}(G)|=1$. The only groups with $|\mathrm{Aut}(G)|=1$ are the trivial group and the group of order $2$ (ref.). The property is true for these two groups.