Suppose that $G$ is a group of order $1536 = 2^9 \cdot 3$.
Question:
How many groups of orders $1536$ are there?
Attempt:
Let $P_2$ and $P_3$ be a Sylow $2$-subgroups and the Sylow $3$-subgroup (there is only one Sylow $3$-subgroup. See Edit 2), respectively. This means that $|P_2| = 2^9$ and $|P_3| = 3$. One can show that $G = P_3 \rtimes P_2$, where $\rtimes$ is the inner semidirect product. Also, if $n_2 = 1$ (i.e., there is only one Sylow $2$-subgroups), then $G \cong P_2 \times P_3$, that is, the direct product of $P_2$ and $P_3$. From this Wikipedia post the maximum number of groups of order $1536$ is bounded above by $1536^{{1536}^{2}}$. However, the same post says that there are exactly $10,494,213$ groups of order $512 = 2^9$ and $1$ group of order $3$ (namely, $\mathbb{Z}_3$). In general, if $N_m$ is the number of groups of order $m = ab < \infty$ with coprime numbers $a$ and $b$, then $N_a \cdot N_b \le N_m \le m^{{m}^{2}}$. Hence, $10,494,213 \le N_{1536} \le 1536^{{1536}^{2}}$.
Restating the question:
What is the exact number for $N_{1536}$, given that $$10,494,213 \le N_{1536} \le 1536^{{1536}^{2}}?$$
EDIT
This question is this paper of Exercises - Group Theory (Exercise 3, pg. 1)
Edit 2
Proof of why $P_3$ (Sylow $3$-subgroup) is the unique normal subgroup of $G$ of order $1536$.
Proof: Recall the following two Sylow Theorems:
Second Sylow Theorem: Let $G$ be a group of order $p^k m$, where $p$ is prime and $p \nmid m$. If $P$ is a Sylow $p$-subgroup and $H$ is another Sylow $p$-subgroup, then they are conjugates. That is, $\exists x \in G$ such that $P = xHx^{-1}$.
Third Sylow Theorem: Let $G$ be a group of order $p^k m$, with $\gcd(p,m)=1$, and $n_p$ denote the number of Sylow $p$ subgroups of $G$. Then, $n_p \equiv 1 \pmod{p}$ and $n_p | m$.
Let $n_p$ denote number of Sylow $p$-subgroups. Since $|G| = 1536 = 2^9 \cdot 3$, let $P_3$ be a Sylow $3$-subgroup of order $3$. Now, let's calculate $n_3$. By the Third Sylow Theorem, $n_3 \equiv 1 \pmod{3}$ and $n_3|2$. Hence, $n_3 \in \{1,4,7,10, \dots\} \cap \{1,2\} = \{1\}$. Therefore, $n_3 =1$.
Now, let's prove normality. By the Second Sylow Theorem, $\exists g \in G$ such that $P_3 = gP_3g^{-1}$. To see this, since $n_3 =1$, then for any other Sylow $3$-subgroup $H$, $$H = P_3$$ However, $H = gP_3g^{-1}$ because both are Sylow $3$ subgroups, and hence, conjugates. Therefore, $$P_3 = gP_3g^{-1}$$ for some $g \in G$.