Let G be a group and $h,g \in G$ with $SgT=ShT$
Show that the subgroups $gTg^{-1}\cap S$ and $hTh^{-1}\cap S$ are conjugated in S.
Two subgroups $U,V$ are conjugated if $\phi(U)=gUg^{-1}=V$, right?
But i don't know how to use this in that case.
Some tips where nice.
You didn't say, but I suppose that $S$ and $T$ are subgroups of $G$.
You need to take an arbitrary element in $gTg^{-1}\cap S$, say $gtg^{-1} = s$, with $s\in S$ and $t\in T$, and show that there is an element $s_{1}\in S$ conjugating it to an element of $hTh^{-1}\cap S$. (And conversely, of course.)
To this end, note that you can write $gt = sg$, so that $g = sgt^{-1}\in SgT = ShT$. Therefore, there are elements $s_{1}\in S$ and $t_{1}\in T$ for which $g = s_{1}ht_{1}$. Now write $$gtg^{-1} = s_{1}ht_{1}t(s_{1}ht_{1})^{-1}.$$ Can you re-arrange this to show that your original element $gtg^{-1}$ is conjugate, by an element of $S$, to some element in $hTh^{-1}\cap S$?
Once you've done that you can get the reverse inclusion either by symmetry or, explicitly, by interchanging the roles of $h$ and $g$.