Given a group $G$ and two conjugated subgroups $ H $ and $ H'=gHg^{-1} $, is the following proposition true?
There are only two possibilities for the subgroups: either $ H\cap H' = 1 $ or $ H=H'$.
I think that these result it's false, but I can't find a counterexample (maybe it's true). Please help me.
On
Actually, there is a name for what you're talking about. Given a subgroup $H\leqslant G$ and a $g\in G$, $H\cap H^g$ is referred to as the twist of $H$ by $g$. (Here $H^g$ denotes $g^{-1}Hg$.) There are many examples where $H\cap H^g$ is nontrivial and proper in $H$, especially amongst Sylow subgroups.
Now, even more relevant to your question is the following definition:
Definition. Let $X\subseteq G$ be a subset. Then $X$ is a T.I. set (short for "trivial intersection set") if for every $g\in G$, either $H\cap H^g=H$ or $H\cap H^g=1$. Of course, if a subgroup happens to be a T.I. set, we call it a T.I. subgroup.
As it turns out, T.I. subgroups are important in representation theory, especially in lemmas used to prove bigger things such as Frobenius's theorem.
One example of where T.I. subgroups naturally arise is in the Sylow $p$-subgroups of the simple groups $PSL(2,p)$ (and, in fact, in any group with self centralizing Sylow $p$-subgroups of prime order that are proper in their normalizers).
Steve D's comment highlights what can happen if the group $G$ has a non-identity normal $2$-subgroup, but $H$ is a Sylow $2$-subgroup which is not normal. Most finite simple groups have a pair of Sylow $2$-subgroups $H$ and $K$ such that $H \cap K \neq 1,$ while $H$ and $K$ are obviously conjugate. An explicit example is the simple group $G = {\rm PSL}(2,11).$