Natural action of $\operatorname{Aut}(G)$ on sets of subgroups of $G$ of same order is transitive. I am looking for the classification of those finite groups whose automorphism group acts transitively on sets of subgroups of same order.
Let $G$ be a finite group and $d$ be a divisor of the order $|G|$ of the group. $A_d$ denote the set of subgroups of $G$ of order $d$. Now Automorphism group $\operatorname{Aut}(G)$ of $G$ acts naturally on $A_d$. This action may or may not be transitive for each divisor $d$ (Leave that $d$ for which $A_d=\emptyset$). 
For exmaple: In Symmetric group $\operatorname{Sym}(n)$ for $(n\geqslant 4)$, above property does not hold but it holds in Alternating group $\operatorname{Alt}(n)$ for $(n\leqslant 5)$.
In particular, If we consider Inner Automorphism group $\operatorname{Inn}(G)$ at the place of $\operatorname{Aut}(G)$, then all B-groups satisfy above property.
Do we already have classification of such groups?
 A: I will call $T$-groups the groups $G$ such that whenever $A_d$ is non-empty, $Aut(G)$ acts transitively on $A_d$. We begin with the $p$-group case, then nilpotent case then...

$\textbf{1}$ Suppose $G$ is an abelian $T$-group (it is also a $p$-group) then $G$ is isomorphic to $\mathbb{Z}_p^n\text{ or }\mathbb{Z}_{p^n} $.

The reason for this is Vanchinathan's comment. Suppose that $G$ is not one of the two above then there exist two groups in $G$, one is isomorphic to $\mathbb{Z}_{p^2}$ (because it is not $\mathbb{Z}_p^n$) and one is isomorphic to $\mathbb{Z}_p\times\mathbb{Z}_p$ (because it is not cyclic). One must use here the classification of abelian group. It thus contains two non-isomorphic groups of order $p^2$.

$\textbf{2}$ Suppose $G$ is a $T$-group and $N$ a characteristic subgroup. Then both $N$ and $G/N$ are $T$-groups.

Because $N$ is characteristic we have two group morphisms from $Aut(G)$ to $Aut(N)$ (by restriction) and $Aut(G/N)$. Then if $H_1$ and $H_2$ are two subgroups of $N$ with $|H_1|=|H_2|$ then there exists $\phi\in Aut(G)$ such that $\phi(H_1)=H_2$. Now $\phi$ induces an automorphism on $N$ so $Aut(N)$ also acts transitively on subgroups of same order. Suppose $Q_1$ and $Q_2$ are subgroups of $G/N$ with $|Q_1|=|Q_2|$ and set $\pi : G\rightarrow G/N$ the canonical projection. Then one can find $\phi\in Aut(G)$ such that $\phi(\pi^{-1}(Q_1))=\pi^{-1}(Q_2)$ because $|\pi^{-1}(Q_1)|=|N||Q_1|=|\pi^{-1}(Q_2)|$, hence $\psi\in Aut(G/N)$ induced by $\phi$ sends $Q_1$ to $Q_2$.

$\textbf{3}$ Suppose $G$ is a non-abelian $T$-group $p$-group then it is a Hamiltonian group.

Let's recall that a Hamiltonian group is a non-abelian group whose every subgroup is a normal subgroup. Take $H$ a subgroup of $G$. Then by property of $p$-groups, one can find a normal subgroup $N$ of order $|H|$. By the $T$-group property  there exists $\phi\in Aut(G)$ such that $\phi(N)=H$, from this it follows that $N\triangleleft G$ implies $H\triangleleft G$.

$\textbf{4}$ Suppose $G$ is a $T$-group, $p$-group then it is either cyclic, $p$-abelian elementary or isomorphic to $\mathbb{Q}_8$. 

If $G$ is abelian then from $\textbf{1}$, we have the first two cases. Suppose that $G$ is not abelian then it is Hamiltonian by $\textbf{3}$ then from a general theorem, it is known that Hamiltonian groups ares isomorphic to $\mathbb{Q}_8\times A\times B$ where $A$ is abelian of exponent $2$ and $B$ is a finite group of odd order (this is definitely a non-trivial result, a proof of it can be found here). In particular, $2$ must divide the order of $G$, hence $p=2$ and hence $B$ is trivial. Then if $A$  were not trivial take $a\in A$ with $a\neq 0$ and $-1\in \mathbb{Q}_8$, I claim that you cannot find an automorphism $\psi$ such that $\psi(-1,0)=(1,a)$. The reason is the following $(i,0)^2=(-1,0)$ but $(0,a)\neq g^2$ for all $g\in G$ (if $g=(s,t)$ then $g^2=(s^2,2t)=(s^2,0)$). Because being a square is an invariant property under automorphism you get that $Aut(G)$ cannot act transitively on subgroups of order $2$ and hence $G$ is not a $T$-group.

$\textbf{5}$ Suppose $G$ is finite nilpotent. Because it is nilpotent, $G$ is a direct product of its $p$-Sylows : $S_{p_1}...S_{p_r}$. Then $G$ is a $T$-group if and only if, for all $i$, $S_{p_i}$ is a $T$-group.

One must use $\textbf{2}$ to show the $\Rightarrow$ case. For the $\Leftarrow$ case, one must use that :
$$Aut(G)=Aut(S_{p_1})\times...\times Aut(S_{p_r})$$
And the fact that every subgroup $H$ is  $\pi_1(H)\times...\times\pi_r(H)$. 
Using $\textbf{4}$ and $\textbf{5}$ we know all the nilpotent finite $T$-groups.
Well, if the $T$-group $G$ is not assumed to be nilpotent it is of course much harder. Considering that groups in the paper you mentioned to us are particular cases of $T$-groups the classification would be far from being trivial. Have you checked for other series of simple groups (Going through the ATLAS of finite groups, you might answer the question in the negative)?
