TITLE: Characterization of finite groups using sum of the orders of their elements
QUESTION [11 upvotes]: Notation: If $G$ is a finite group, $o(g)$ denotes the order of the element $g\in G$. 
Motivation: Some finite groups could be uniquely determined by the size of the group. For example given a prime number $p$ there is just one group $\mathbb{Z}_p$ up to isomorphism. As there are infinitely many prime numbers, the number of finite groups that are completely determined with their size is infinite.  
On the other hand there are finite groups that could be uniquely determined by the sum of the orders of their elements. For example $A_5$ is the unique finite group $G$ with $\Sigma_{g\in G}o(g)=211$ (see this paper by Habib Amiri, S. M. Jafarian Amiri & I. M. Isaacs). 
Inspired by this fact the following question arises:  

Question: For which natural numbers like $n$, there is a unique finite group up to isomorphism with $n=\Sigma_{g\in G}o(g)$? Is there infinitely many such natural numbers?

REPLY [5 votes]: As a counterbalance to the question,  note  that for any odd prime $p$, there are two non-isomorphic groups of ordér $p^{3}$ with sum of element orders $p^{4} -p+1$.
Any $p$-group with that element order sum has order at most $p^{3}$. Groups of exponent $p$ are dealt with, and a $p$-group with more than one cyclic subgroup of order $p^{2}$ has element order sum too large- likewise for a $p$-group of order $p^{3}$ with a unique subgroup of order $p^{2}$. Cyclic $p$-groups are easily dealt with. So there are no more $p$-groups with that element order sum.