Solution:

Let $P_{1}$ and $P_{2}$ be the two congruent pentagons. Let $p(P)$ denote the perimeter of polygon $P$.
We give an upper bound for $p(P_{1}) + p(P_{2})$. Note that since a square has four sides, at least four sides of $P_{1}$ and $P_{2}$ combined lie on the sides of the square. These sides have total length at most $4$, the perimeter of $ABCD$.
Each of the remaining sides has length at most $\sqrt{2}$, since the longest possible length of a segment inside $ABCD$ is $\sqrt{2}$. There are at most $6$ remaining sides, so
$$p(P_{1}) + p(P_{2}) \leq 4 + 6\sqrt{2}.$$
Since $P_{1}$ and $P_{2}$ are congruent, this implies
$$p(P_{1}) = p(P_{2}) \leq \boxed{2 + 3\sqrt{2}}.$$
This least upper bound can be achieved by placing $X$ close to $C$ and $Y$ close to $A$, as seen in the diagram.