Finite abelian groups with fewer automorphisms than a subgroup It is not too hard to find examples of finite groups which have fewer automorphisms than one of their subgroups. For example $\mathcal D_4 \times \mathbb Z/2\mathbb Z$ (where $\mathcal D_4$ is the dihedral group of order $8$) has at most (in fact exactly) $64$ automorphisms but contains a subgroup isomorphic to $\left(\mathbb Z/2\mathbb Z\right)^3$, which has $\operatorname{GL}_3(\mathbb F_2)$ as automorphism group, with $168$ elements.
Every example I found was non-abelian, so I wondered if a finite abelian group can have less automorphism than one of its subgroup? Hillar and Rhea - Automorphisms of finite Abelian groups gives the number of automorphisms of an abelian $p$-group, but it's not clear to me if that function is non-decreasing with cardinality.
 A: From the Hiller-Rhea formula
$$|\operatorname{Aut} H_p| = \prod_k (p^{d_k} - p^{k-1}) \prod_j (p^{e_j})^{n-d_j} \prod_i (p^{e_i-1})^{n-c_i+1},$$
given an abelian $p$-group of type $p^{e_1}\cdots p^{e_n}$ where $1 \leq e_1 \leq \cdots \leq e_n$, you can try to increment one of the exponents $e_l$ by 1, and see how things change.  Here, $d_k = \max \{l | e_l = e_k\}$ and $c_k = \min \{ l | e_l = e_k \}$.
Case 1: $e_{l-1} < e_l$ and $e_l + 1 < e_{l+1}$. The only changes to the formula are $(p^{e_l})^{n-d_l}$ and $(p^{e_l-1})^{n-c_l+1}$, and yield a strict increase.
Case 2: $e_l$ is isolated, but $e_l + 1$ is equal to $m$ other values.  Then, $d_l$ jumps by $m$, and $c_i$ decrease by $1$ for $m$ entries.  We end up adding an exponent of $2n - 2l + 1 + \epsilon > 0$, where $\epsilon$ is a way to ignore the $p^{k-1}$ in the first product.
Case 3: $e_l$ is equal to $m$ other values, but $e_l+1$ is isolated.  Then, $d_i$ drops by $1$ for $m$ entries, and $c_l$ jumps by $m$.  We add an exponent of $2n-2l+1 - \epsilon > 0$.
Case 4: $e_l$ is equal to $m$ other values, and $e_l+1$ is equal to $r$ other values.  Then, $d_i$ drops by $1$ for $m$ entries, $d_l$ jumps by $r$, $c_i$ decrease by $1$ for $r$ entries, and $c_l$ jumps by $m$.  We add an exponent of $2n-2l+1 - \epsilon > 0$.
In conclusion, a subgroup of a finite abelian group always has strictly smaller automorphism group (Edit: with the exception of a group of order $2$ times an odd number, and the subgroup of index 2, where we have equality).  The minimal case is when $e_1 = \cdots = e_n$.  Then, incrementing $e_n$ multiplies the order by $\frac{p^n(p-1)}{p^n-1}$, which approaches 1 when $p=2$ and $n$ is large.
