The question is in the title.
My attempt:
Since $\Vert F(x) \Vert = \Vert F^*(x) \Vert, \forall x \in \mathbb{C^n}$, then $\Vert F(x) \Vert^2 = \Vert F^*(x) \Vert^2, \forall x \in \mathbb{C^n}$. So, $\forall x \in \mathbb{C^n}$, we have: $$\Vert F(x) \Vert^2 = \langle F(x), F(x) \rangle = \langle F^*F(x), x \rangle$$ $$\Vert F^*(x) \Vert^2 = \langle F^*(x), F^*(x) \rangle = \langle FF^*(x), x \rangle$$ And we can conclude from the equality given that: $$\forall x \in \mathbb{C^n}: F^*F(x) = FF^*(x) \implies FF^* = F^*F$$
I think my argument is false, or at least insufficient. How can i improve it?
You have proved that $\langle FF^*(x), x \rangle=\langle F^*F(x), x \rangle$ or $\langle (FF^*-F^*F)(x), x \rangle=0$ for all $x$. This implies that $\langle (FF^*-F^*F)(x), y \rangle=0$ for all $x$ and $y$ and you can take $y=(FF^*-F^*F)(x)$ to finish the proof. [Details below].
Supppose $\langle Tx, x \rangle=0$ for all $x$. Using the equations $\langle T(x+y), x+y \rangle=0$ and $\langle T(x-y), x-y \rangle=0$ conclude that $$\langle Tx, y \rangle=-\langle y, Tx \rangle.$$ Now use the equations $\langle T(x+iy), x+iy \rangle=0$ and $\langle T(x-iy), x-iy \rangle=0$ conclude that $$\langle Tx, y \rangle=\langle y, Tx \rangle.$$ Combining the two we get $\langle Tx, y \rangle=0$.