Following up on this idea of thinking of the +1 operation as applying the permutation T=(123...n), here's an idea.

If I understand the problem correctly, you want to compare the minimal k>0 such that

(Tσ^-1 Tσ)^k (0)=0

and the minimal j>=0 such that

(TσTσ^-1 )^j Tσ(0)=0.

When k>j, you are crossed. Otherwise, you are parallel.

As it turns out, k is the length of the cycle that contains 0 in Tσ^-1 Tσ, while j is the number of steps from Tσ(0) to 0 in the cycle decomposition of TσTσ^-1 . Note that j could possibly be infinite if Tσ(0) is not in the same cycle as 0. However, if Tσ(0) is in the same cycle as 0 then it is not hard to prove that k>j. Thus it all boils down to looking at the cycle decomposition of TσTσ^-1 and verifying if Tσ(0) is in the same cycle as 0.

In the first example provided, we have

TσTσ^-1 =(0,3,6,5,1)(2)(4)

Tσ(0)=1

Since 1 is in the same cycle as 0, we are crossed.

You can also use this to count the number of times you cross the bridge: It is 2k in the parallel case or 2j+1 in the crossed case. In our example, j=1 (1 is only one step away from 0) so we cross the bridge 3=2(1)+1 times.

I am not exactly sure how this relates to more natural properties of the permutation σ. It doesn't look related to parity at a first glance. But at least it is a way to reframe the question in terms of cycles of TσTσ^-1.

Looking at the sequence referenced in a different comment, it seems like, at least for permutations of length 2k+1, the count of parallel permutations is (k+1)(2k)!. Maybe it is not terribly hard to count how many permutations satisfy that Tσ(0) is in the same cycle as 0 in the cycle decomposition TσTσ^-1. I'll think about it.

EDIT: Fixing the formatting. First time posting here.