Ship lines maintained by Red dragon we will call red lines, and those maintained by Blue eye blue lines.

Let $a_n$ denote the number of good arrangements for an archipelago with $n$ islands. Consider an archipelago with $n + 1$ islands and notice island $1$. Let $A$ be the set of all islands which are connected to island $1$ with a red line, and let $B$ be the set of all islands which are connected to island $1$ with a blue line.

Note that all lines (if there are any) from some island in $A$ to some island in $B$ must be blue. Suppose that's not the case, i.e. that there are islands $a \in A$ and $b \in B$ such that $a < b$ and the line $a \to b$ is red. We know that line $1 \to b$ is blue, and lines $1 \to a$ and $a \to b$ are red, hence we can go from $a$ to $b$ using only red lines, and also using only blue lines, which is a contradiction. By the same reasoning we conclude that all lines (if there are any) from some island in $B$ to some island in $A$ must be red.

Let set $A$ contain $k \in \{0, 1, \dots, n\}$ islands. That means that set $B$ contains $n-k$ islands. Note that islands in the set $A$ must form an archipelago with a good arrangement of ship lines, and the same holds for the set $B$. On the other hand, for each selection of archipelagos $A$ and $B$ with good arrangements we get exactly one archipelago with $n+1$ islands and good ship lines arrangement between them.

Now, using mathematical induction, we will prove that $a_n = n!$. It is obvious that $a_0 = a_1 = 1$. Suppose that $a_k = k!$, for all $k \le n$, for some positive integer $n$. By previous reasoning we conclude that
$$a_{n+1} = \sum_{k=0}^{n} \binom{n}{k} a_k a_{n-k} = \sum_{k=0}^{n} n! = (n+1)!$$
Thus, $a_n = n!$ and finally $a_{2017} = 2017!$.