WLOG the crossing starts clockwise and then multiply by two. Define a returning island as an island $ i$ such we go from island $ i$ to island $ i + 1$ and returning back to island $ i$. (Same with $ n$ to $ 1$ and back to $ n$). Case 1: Path with no returning island, $ 123 \cdots n123\dots n1$ there $ 2n$ steps to take with two options on each case then $ 2^{n}$ crossings Case 2: Path with one returning island $ i$, $ 12 \cdots i \cdots 21 n(n - 1) \cdots (i + 1)i(i + 1) \cdots (n - 1)n1$ then $ 2^{n} \times n$ crossings. Case 3: Path with $ 2$ or more returning islands, no possibilities. To sum up total crossings $ = 2 \times (2^{n} + 2^{n} \times n) = (n + 1) 2^{n + 1}$ Edited with correction...from next post.

I think there is a mistake in your solution... Where it is $ 2n$ it should be $ n$... For example, in case 1, on the route $ X_1 - X_2 - \cdots - X_n - X_1$, there are $ 2^n$ possibilities, because from one island to another there are always two options. But when we do the same route again, from one island to another, we can't pass through one of the bridges, because we have already passed through it, so there is only one possibility... In case 2 you have the same mistake. Regardless of that, my solution is similar to yours...

But... I have a question, can you return?, I say if it is possible that you can go from X_1 to X_2, and then return from X_2 to X_1? I am waiting for an answer.

First of all, where it is written $ n\ge2$ it should be written $ n>2$. In Portuguese it's that. Anyway, it doesn't matter. Now answering to the question... Yes, it is. But you can only return once because otherwise you won't be able to do the entire route as you want. I think it's easy to see that...

Own correction: when you return once you return twice, even though the returning island is the same, as shown by mszew.

I claim that there are $\boxed{n \cdot 2^{n + 1}}$ total sequences. Let $a_n$ be the number of distinct bridge sequences. Then we prove the following by induction: Claim: $a_n = n \cdot 2^{n + 1}$, for $n \geq 3$. Proof: The base case for $n = 3$ is trivial. Assume the statement is true for all $k \leq n$, and now we prove for the $(n + 1)$-st case: Consider player $X_{n + 1}$, and observe that for every path consisting of $n$ people, we must pass through $X_{n + 1}$ and further every path through $X_{n + 1}$ has an additional two bridge options. (Also observe that the symmetry in whichever way we began with - clockwise or counterclockwise - is accounted for, as in this case every path is dependent on the paths from $a_n$). Now we have yet to count the fully clockwise and fully counterclockwise paths with $X_{n + 1}$ as an island, and so accounting for these final paths, one obtains; \[a_{n + 1} = 2a_n + 2(2^{n + 1}) = 2a_n + 2^{n + 2} = (n + 1) \cdot 2^{n + 2}, \]as claimed. $\blacksquare$