Let $\phi:G \to G$ be the automorphism. We'll prove $\phi(P)$ is a $p$-Sylow Subgroup of $G$.
$\phi$ takes identity to itself:
For any $e_G \in G$, the identity in G, $\phi(e_G.e_G)=\phi(e_G)\phi(e_G)$
which proves the result.(from cancellation law in a group)
$\phi$ takes inverses to inverses:
Use the fact that $gg^{-1}=e_G$ for any $g \in G$. Do a computation similar to the one above.
$\phi$ takes subgroups to subgroups:
Let $H \leq G$. We intend to prove $\phi(H)$ is a subgroup. Use the 'lemma' we have proved before and verify the subgroup criterion (that $\phi(H)$ is closed under multiplication and inverses. )
Now, by one of my comments above, (in fact by just using the bijectivity of the map $\phi$, and by looking at its restriction to $H$), we'll prove that $|\phi(H)|=|H|$.
Note that the definition for two sets to be of same cardinality is that there exists a bijection between them.
So, You are through.
Best Answer
Examine the action of $G$ on the set of Sylow 3-subgroups. This gives rise to a well-defined homomorphism $\phi: G\to S_4$. Since we know the Sylow 3-subgroups are all conjugate to one another, the image $\phi(G)$ must be a subgroup of $S_4$ that acts transitively on the set with four points. The only such subgroups of $S_4$ that could arise as a quotient a group of order 36 are isomorphic to either the cyclic group of order 4, the Klein 4 group, or the alternating group $A_4$.
Note that if $|\phi(G)| = 4$, then the kernel of $\phi(G)$ is a normal subgroup of order 9, contradicting our assumption that the Sylow 3-subgroups are not normal.
Now if $\phi(G)$ is isomorphic to $A_4$, then $\ker \phi$ is a normal subgroup of order 3. This implies that $\ker \phi$ is a subgroup of all Sylow 3-subgroups. Since all groups of order 9 are abelian, this means that the centralizer of $\ker \phi$ is at least order 27 (six unique elements from each of the four Sylow 3-subgroups as well as $\ker \phi$ itself) and thus $\ker \phi$ is actually central.
Let $x$ denote a nonidentity element of $\ker \phi$ and $y$ denote an element of order dividing 4. The subgroup generated by $y$ has trivial intersection with $\ker \phi$, so $y$ must actually have order $1$ or $2$, since the image $\phi(G)$ is a group with no elements of order 4.
Moreover, $(yx)^2 = y^2x^2 = x^2$, showing that $yx$ does not have order 2. Thus, the left cosets of $\ker \phi$ contain at most one element of order dividing 2. Since there are only four cosets that could contain such an element (as only four elements in the quotient group have order dividing 2), there are at most 4 such elements, meaning that the Sylow 2-subgroup is unique.