All the edges are \(1\) unit.

Let \(\overline{DF}=x\) unit.

\(\therefore\) \(\overline{C^{\prime}F}=1-x\)

As, \(\angle C^{\prime}DF=90^o\) we can apply the \(\textbf{Pythagorean Theorem}\) on that triangle,

\({\frac{1}{3}}^2+x^2=(1-x)^2\)

\(\Rightarrow x=\frac{4}{9}\)

\(\therefore\) \(FC^{\prime}=\frac{5}{9}\)

Now, if the \(\angle FC^{\prime}D\) be \(\theta\) then \(\angle EC^{\prime}A=90^o-\theta\).

So, \(\triangle AEC^{\prime}\sim\triangle DFC^{\prime}\)

\(\therefore\) \(\frac{DF}{AC^{\prime}}=\frac{C^{\prime}D}{AE}=\frac{C^{\prime}F}{C^{\prime}E}\)

From, \(\frac{DF}{AC^{\prime}}=\frac{\frac{4}{9}}{\frac{2}{3}}=\frac{2}{3}\)

So, \(\frac{C^{\prime}D}{AE}=\frac{2}{3}\Rightarrow AE=\frac{1}{3}\times\frac{3}{2}=\frac{1}{2}\)

\(\frac{C^{\prime}F}{C^{\prime}E}=\frac{2}{3}\Rightarrow C^{\prime}E =\frac{5}{9}\times\frac{3}{2}=\frac{5}{6}\).

So the perimeter of the \(\triangle AEC^{\prime}=\frac{2}{3}+\frac{1}{2}+\frac{5}{6}=\boxed{2}\).