Because $ABCD$ is a square, point $D$ will move to point $B$. Let the new location of $F$ be $F'$. It must be true that $\angle{FAF'} =90^\circ$. Because $\triangle{AFD} \cong \triangle{AF'B}\implies AF=AF'$, we find $\triangle{F'AF}$ is an isosceles right triangle. $$\therefore\quad\angle{AFF'} = 45^\circ \quad \text{and}\quad FF' = \sqrt{2^2+2^2}=2\sqrt{2}$$ Now the three sides of $\triangle{F'FB}$ are $2\sqrt{2}$, 1 and 3. By Pythagorean theorem, it is a right triangle with $\angle{F'FB}=90^\circ$. It follows that $\angle{AFB} = \angle{AFF'} + \angle{F'FB} = 45^\circ + 90^\circ = \boxed{135^\circ}$.