Let $G$ be a finite group and $p$ a prime such that $p^\alpha$ divides $|G|$ and $p^{\alpha+1} \nmid |G|$. I know that Sylow $p$-subgroups of $G$ are conjugate to one another but if we have some $p$-subgroups of $G$ with the same cardinality(different from $p^\alpha$) and each of them are isomorphic to some Sylow $p$-subgroup(the Sylow p subgroup can be the same or different) of some other group(group has to be the same)...can we say that they are conjugate in $G$?
I can't think of a good example, but this problem below lead me to the question above: In $S_4$ with $|S_4|=2^33$we have $\langle (1234)\rangle$, $\langle (1243)\rangle$, $\langle (1324)\rangle$ as $2$-subgroups of $G$ isomorphic to $\mathbb{Z} _4$. A Sylow $2$-subgroup in $\mathbb{Z} _4$ has cardinality $2^2$. So the images of $\langle (1234)\rangle$, $\langle (1243)\rangle$, $\langle (1324)\rangle$ are conjugate in $\mathbb{Z} _4$. Can I say that $\langle (1234)\rangle$, $\langle (1243)\rangle$, $\langle (1324)\rangle$ are Sylow $2$-subgroupsconjugate in $S_4$
This is a bad example but do you get what I'm trying to say?
Here's a negative example. Take two groups of order 2 in the dihedral group of 4 elements $D_{2\cdot2}\cong V_4$: $H=\{e,\rho\}$ and $K=\{e,\tau\}$. These are normal in $D_{2\cdot2}$ (index 2) so aren't conjugate, but are isomorphic to Sylow-2 groups in the dihedral group $D_{2\cdot3}$ (pick two reflection subgroups of order 2). So as darij said a semidirect product will work.