In this answer, I tried to address all the OP' questions from the post and comments and realize their idea.
Given a homomorphism $\varphi:G/\mathcal Z(G)\to G,$ with the property $\varphi(g\mathcal Z(G))\in g\mathcal Z(G),$ we can conclude the following:
- $\varphi$ is injective:
indeed, if $g_1\mathcal Z(G),g_2\mathcal Z(G)\in G/\mathcal Z(G),g_1\mathcal Z(G)\ne g_2\mathcal Z(G),$ then $g_1\mathcal Z(G)\cap g_2\mathcal Z(G)=\emptyset$ and $\varphi(g_1\mathcal Z(G))\in g_1\mathcal Z(G)$ and $\varphi(g_2\mathcal Z(G))\in g_2\mathcal Z(G)$ must be different as elements of two disjoint sets.
Since $\varphi$ is injective, $G/\mathcal Z(G)\cong\operatorname{im}\varphi.$
$\operatorname{im}\varphi$ intersects each coset $g\mathcal Z(G)$ at exactly one point.
Otherwise, if there were $g\in G$ such that $\operatorname{im}\varphi\cap g\mathcal Z(G)$ contained two distinct elements $$h_1, h_2,\in g\mathcal Z(G),h_1=\varphi(g_1\mathcal Z(G)), h_2=\varphi(g_2\mathcal Z(G)),$$ then $g_1\mathcal Z(G),g_2\mathcal Z(G)\in G/\mathcal Z(G)$ would have been distinct with $$\varphi(g_1\mathcal Z(G))\mathcal Z(G)=h_1\mathcal Z(G)=g\mathcal Z(G)=h_2\mathcal Z(G)=\varphi(g_2\mathcal Z(G))\mathcal Z(G),$$ (because $h_i\in g\mathcal Z(G)\Leftrightarrow h_i\mathcal Z(G)=g\mathcal Z(G)$) and we reach a contradiction with 1.
- $\operatorname{im}\varphi\cap\mathcal Z(G)=\{e_G\}.(*)$
From the previous point, having in mind $\varphi(\mathcal Z(G))=e_G\in\mathcal Z(G)\le G,$ we conclude that $\operatorname{im}\varphi\cap\mathcal Z(G)=\{e_G\}.$
Now, $$\begin{aligned}&\forall g\in G,\color{blue}{\varphi(g\mathcal Z(G))\in} g\mathcal Z(g)\overset{\substack{\mathcal Z(G)\\\text{normal}\\\text{ in } G}}{=}\color{blue}{\mathcal Z(G)g}\\\Leftrightarrow &\forall g\in G, g\in\mathcal Z(G)\varphi(g\mathcal Z(G))\\\Leftrightarrow &\forall g\in G,\exists z(g)\in\mathcal Z(G),\boxed{g=z(g)\varphi(g\mathcal Z(G))}.(**)\end{aligned}$$
$\operatorname{im}\varphi$ is normal in $G.$
Let $\varphi(g\mathcal Z(G))\in\operatorname{im}\varphi$ and $x\in G$ be arbitrary. Then, by the previous point: $$\begin{aligned}x\varphi(g\mathcal Z(G))x^{-1}&=\underbrace{z(x)}_{\in\mathcal Z(G)}\varphi(x\mathcal Z(G))\varphi(g\mathcal Z(G))\varphi(x\mathcal Z(G))^{-1}\underbrace{z(x)^{-1}}_{\in\mathcal Z(G)}\\&\overset{\substack{\varphi \text{ is a}\\\text{homo-}\\\text{ morphism}}}{=}z(x)z(x)^{-1}\varphi(xgx^{-1}\mathcal Z(G))\\&=\varphi(xgx^{-1}\mathcal Z(G))\in\operatorname{im}\varphi\end{aligned}$$
Therefore:
(1) $\operatorname{im}\varphi,\mathcal Z(G)\unlhd G$
(2) $\operatorname{im}\varphi\cap\mathcal Z(G)=\{e_G\}$ (by $(*)$).
(3) $G\overset{(**)}{=}\operatorname{im}\varphi\mathcal Z(G)$
and the claim follows just as the OP had in mind.
Here’s a more conceptual solution which I think is cleaner. How many times is $7!$ divisible by $2$? A quick check yields $2^4$, so a $2$-Sylow subgroup of $S_7$ has order $16$.
Remember that all $2$-Sylow subgroups are conjugate, hence isomorphic. Your task, then, is to find such a group, and show that it is not isomorphic to $C_4\times C_4$, the product of two cyclic groups.
Constructing a group of order $16$ inside $S_7$ isn’t very hard: you should be familiar by now with dihedral groups. A dihedral group of size $8$ is easy to spot inside $S_7$, and then just multiply it by a cyclic group of order $2$ to obtain a Sylow subgroup. Finally, observe that this group isn’t isomorphic to $C_4\times C_4$ by counting elements of various orders (or, if you’re less silly than I am, by observing that one of these groups is abelian and the other is not.)
Best Answer
To show $\phi $ is a homomorphism, we need to prove that $\phi ((h_1,n_1)\cdot (h_2,n_2))=\phi ((h_1,n_1))\phi ((h_2,n_2)). $
Note that $$\phi ((h_1,n_1)\cdot (h_2,n_2))=\phi((h_1h_2,n_1n_2))=h_1h_2n_1n_2=(h_1n_1)(h_2n_2)\\=\phi ((h_1,n_1))\phi ((h_2,n_2)).$$