The second to last line is where you reach a problem: We don't have an "image" of $S_3$, we have a pre-image. This pre-image contains $P_3$, but is not in necessity equal to it. In fact, since the Slyow 3-group of $S_3$ contains the identity, the pre-image of the subgroup of $S_3$ contains the kernel, properly. But the pre-image of a normal subgroup is normal. So we must only really show that the pre-image is a proper subgroup.
A simpler proof though, notes that we may let $\phi$ be the representation by conjugation on the Slyow 2-groups (which are of order 16). If this is trivial, then we see that $n_1=1$ by Slyow's Theoroms. If not, then Ker $\phi$ is the desired proper normal subgroup (Since Ker $\phi=0$ makes no sense considering the order).
$\mathbb{Z}_5\times \mathbb{Z}_5$ can be considered as a vector space over field $\mathbb{Z}_5$ (the addition is same as addition in abelian group modulo $5$, and scalar multiplication actually comes from addition: $v+v=2.v$, $v+v+v=3.v$, $\cdots$)
Thus any automorphism of the group $\mathbb{Z}_5\times \mathbb{Z}_5$ is also an automorphism of the vector space $\mathbb{Z}_5\times \mathbb{Z}_5$ and conversely (because of remark on scalar multiplication made earlier).
It is well known, what is the automorphism group of $\mathbb{Z}_5\times \mathbb{Z}_5$? It is
$$
GL(2,5)=
\begin{Bmatrix}
\begin{bmatrix}
a & b\\
c & d
\end{bmatrix}\colon a,b,c,d\in\mathbb{Z}_5, ad-bc\neq 0
\end{Bmatrix}.
$$
To get an element of order $3$ in $Aut(\mathbb{Z}_5\times \mathbb{Z}_5)$, we have to find a non-identity matrix $A$ such that $A^3=I$. Thus, $A$ satisfies polynomial $x^3-1=(x-1)(x^2+x+1)$. The quadratic factor has no root in $\mathbb{Z}_5$ (check) so it is irreducible. Further, $A\neq I$, hence the minimal polynomial of $A$ should divide quadratic factor $x^2+x+1$. But the minimal polynomial of $A$ has degree $\leq 2$ (size of $A$), so $x^2+x+1$ is the minimal (hence characteristic polynomial of $A$). Can we find a matrix $A$ explicitly with such characteristic polynomial? Yes; companion matrix
$$
A=
\begin{bmatrix}
0 & -1\\
1 & -1
\end{bmatrix}.
$$
Now we come to construction of group. Let $\mathbb{Z}_5\times \mathbb{Z}_5=\langle x,y\rangle$ with $x=\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $y=\begin{bmatrix} 0 \\ 1 \end{bmatrix}$. Then $Ax=y$ and $Ay=-x-y$ (check).
Consider $\mathbb{Z}_5\times \mathbb{Z}_5=\langle x,y\rangle$ multiplicatively, $t$ be element of order $3$ in its automorphism group, and action of $t$ on this group is given by (write above action of $A$ as action of $t$, with multiplicative operation):
$$txt^{-1}=y \,\,\,\, tyt^{-1}=x^{-1}y^{-1}.$$
Thus
$$G=\langle x,y,t\colon x^5=y^5=1, xy=yx, t^3=1, txt^{-1}=y, tyt^{-1}=x^{-1}y^{-1}\rangle$$
is a group of order $75$, it is non-abelian.
Best Answer
Sylow theory is generally not useful for Abelian groups since we already know so much about their structure, and since every Abelian group has a unique $p$-Sylow subgroup when it exists.
Using the classification theorem, we have that $|{G}| = 75$ implies $G \cong \mathbb{Z}/3\mathbb{Z} \times (\mathbb{Z}/5\mathbb{Z})^2$ or $G \cong \mathbb{Z}/3\mathbb{Z} \times \mathbb{Z}/25\mathbb{Z} \cong \mathbb{Z}/75 \mathbb{Z}$.
Now we use standard combinatorical methods to count the automorphisms in each case.
In the first case we must send $(1,0,0)$ to an element of order $3$, giving $2$ choices. We must send $(0,1,0)$ to an element of order $5$, giving $24$ choices. We must send $(0,0,1)$ to an element of order $5$ so that our function is surjective, giving $24 - 4 = 20$ choices. That is we are subtracting off the elements in the subgroup generated by the image of $(0,1,0)$. After defining our function on a basis, we can extend uniquely to a homomorphism on the group. This gives $20 \cdot 24 \cdot 2 = 960$ automorphisms. This work is made easier by noting $\text{aut}(H) \times \text{aut}(K) \cong \text{aut}(H \times K)$ when $H$ and $K$ are abelian groups of relatively prime orders.
In the second case we must simply pick a generator, giving $\phi(75)$ automorphisms, where $\phi$ is the Euler totient function.