Groups having at most one subgroup of any given finite index Cyclic groups have at most one subgroup of any given finite index. Can we describe the class of all groups having such property?
Thank you!

Let $G$ be a group.  The canonical residually finite quotient of $G$ is $R(G)=G/K$ where $K$ is the intersection of all the finite-index subgroups of $G$.
Lemma: If $G$ is finitely generated (update) then $G$ has at most one subgroup of each index if and only if $R(G)$ is cyclic.
Proof: First, note that $R(G)$ is residually finite.  If every finite quotient of $R(G)$ is cyclic then $R(G)$ is residually cyclic, and it follows that $R(G)$ is abelian.  So $R(G)$ has a non-cyclic finite quotient unless $R(G)$ is cyclic.  Therefore, if $R(G)$ is not cyclic then $R(G)$, and hence $G$, has a finite non-cyclic quotient, and hence, by Artuto's answer, has a two distinct finite-index subgroups of the same index.
Conversely, suppose that $R(G)$ is cyclic.  Every finite-index subgroup of $G$ contains $K$, so the quotient map $G\to R(G)$ maps finite-index subgroups to finite-index subgroups bijectively and preserves the index.  Therefore, if $R(G)$ is cyclic then $G$ has at most one subgroup of each index.  QED
I believe that it is an open question whether or not there is an algorithm to determine whether a fp group has a proper finite-index subgroup, ie whether or not $R(G)$ is non-trivial.  So it may be open whether or not it is possible to determine if $R(G)$ is cyclic, too.
Note: Earlier, I forgot to mention that I had implicitly assumed that $G$ is finitely generated.  This assumption is clearly necessary; otherwise the additive group of the rationals is a counterexample.  If $G$ is not finitely generated, then the same argument shows that if $G$ has at most one subgroup of each finite index then $R(G)$ is residually cyclic.  But it's not clear to me that the converse of this statement is true.   So I'll finish with a question:

If $G$ is residually cyclic, does $G$ have at most one subgroup of each finite index?

