Rotate $\triangle{CNB}$ around point $C$ for $90^\circ$ clockwise. Because $\triangle{ACB}$ is an isosceles right triangle, point $B$ will move to point $A$. Let the new location of $N$ be $N'$.

Note that $\angle{N'AM}=\angle{N'AC}+\angle{CAM}=\angle{NBC}+\angle{CAM}=90^\circ$. Therefore, by Pythagorean theorem, we find $$MN'=\sqrt{N'A^2 + AM^2}=\sqrt{NB^2 + AM^2}=5$$ Meanwhile, because $\angle{N'CM} = \angle{N'CA} +\angle{ACM} = \angle{BCN} + \angle{ACM} = \angle{MCN}$ and $CN'=CN$, we find $\triangle{CN'M} \cong \triangle{CNM}$ by SAS. Hence $$MN=MN'=\boxed{5}$$