Here's the complete problem.
Let $G$ be group and $f:G\rightarrow G$ be automorphism. Define a set $Z=\lbrace z \in G | zg=gz, \forall g \in G \rbrace$ and $f(Z)=\lbrace F(z)|z \in Z \rbrace.$ Prove that $f(z)=Z$.
Here I tried.
Firstly, did the notation $F(z)$ on $f(Z)=\lbrace F(z)| z \in Z \rbrace$ should be change by $f(z)$ and on $f(z)=Z$, $f(z)$ should be change by $f(Z)$? If, no, I can't solve it.
If yes, I can solve it and here it is.
We'll show that $f(Z)=Z$.
Let $x \in f(Z)$. Then, $f^{-1}(x) = Z$ such that $f^{-1}(x)g=gf^{-1}(x)$. Since $f(f^{-1}(x)g)=xf(g)=f(g)x=f(gf^{-1}(x)$, then $x \in Z$. Hence, $f(Z) \subseteq Z$.
Conversely, let $y \in Z$ such that $yg=gy$. Since $f^{-1}(yg)=f^{-1}(y)f^{-1}(g) = f^{-1}(g)f^{-1}(y)$
$= f^{-1}(gy)$, then $f^{-1}(y) \in Z$ and implies that $y \in f(Z)$. Hence, $Z \subseteq f(Z)$.
Hence, $f(Z)=Z$.
Thanks for correction and help in advanced.
You're proving the center is characteristic.
This should be pretty straight forward: $x\in Z\iff xg=gx,\,\forall g\iff f(g)f(x)=f(gx)=f(xg)=f(x)f(g)\iff f(x)\in Z$.
I couldn't quite follow your proof. But the change from $z$ to $x$ is probably a good move.