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](https://arxiv.org/abs/math/0605185) 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.



Finite abelian groups with fewer automorphisms than a subgroup
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.


