imc

We can set the point on $CD$ where the fold occurs as point $F$. Then, we can set $FD$ as $x$, and $CF$ as $1-x$ because of symmetry due to the fold. It can be recognized that this is a right triangle, and solving for $x$, we get, $$x^2+{\left(\frac{1}{3}\right)}^2=(1-x{)}^2\to x^2+\frac{1}{9}=x^2-2x+1\to x=\frac{4}{9}$$ We know this is a 3-4-5 triangle because the side lengths are $\frac{3}{9},\frac{4}{9},\frac{5}{9}$. We also know that $EA{C}^{\prime }$ is similar to $C^{\prime }DF$ because angle $E{C}^{\prime }F$ is a right angle. Now, we can use similarity to find out that the perimeter is just the perimeter of $C^{\prime }DF\times \frac{A{C}^{\prime }}{DF}$. That's just $\frac{4}{3}\times \frac{\frac{2}{3}}{\frac{4}{9}}=\frac{4}{3}\times \frac{3}{2}=2$. Therefore, the final answer is $\boxed{2}$