I think what you are looking at is a pivot vector in "Nag Form". There are algorithms to transform it into the standard form of a permutation matrix, but the algorithm isn't clear to me.
Here are some references to routines that refer to the format:
Maple - Create Permutation
Calling NAG Routines from Matlab
The group of permutation matrices of size $n$ is isomorphic to the symmetric group $S_n$. Any permutation $\sigma$ has finite order, hence any permutation matrix also has finite order.
Added: decomposition of a permutation as a product of disjoint cycles (example)
Let's take the permutation of $S_8$:
$$\sigma=\begin{pmatrix}1&2&3&4&5&6&7&8\\5&4&6&3&1&7&8&2\end{pmatrix}$$
We also can represent this permutation in the following way: starting with $1$, we compute its image by $\sigma$, then the image of $\sigma(1)$, and so on, until we get back to $1$. When this is done, we have obtained a cycle. If there remains elements in $\{1,2,\dots,8\}$, we do the same starting from the smallest element.
Here we obtain:
$$\sigma=(1\,5)(2\,4\,3\,6\,7\,8)$$
Thus $\sigma$ is the product of a $2$-cycle (a ‘transposition’) and a $6$-cycle.
Now a cycle of length $\ell$ has order $\ell$: indeed
\begin{align*}
\sigma^2&=(2\,3\,7)(4\,6\,8)& \sigma^3&=(2\,6)(3\,8)(4\,7)\\
\sigma^4&=(2\,7)(3)(4\,8)(6)& \sigma^5&=(2\,8\,7\,6\,3\,4) \\
\sigma^6&=\operatorname{id}
\end{align*}
Also, the order of the product of disjoint cycles is the l.c.m. of the orders (lengths) of the cycles.
Best Answer
The "permutation matrix" associated to $\pi$ is the matrix that is obtained from the identity matrix by "swapping columns" according to the permutation $\pi$.
For example, if $$\pi = \left(\begin{array}{cccc} 1 & 2 & 3 & 4\\ 2 & 3 & 1 & 4 \end{array}\right),$$ then the permutation matrix would be the matrix obtained from the identity by moving the first column to the 2nd column position; the second column to the third column position; the third column to the first column position; and leaving the fourth column in the fourth column position. That is, $$P_{\pi}=\left(\begin{array}{cccc} 0 & 1 & 0 & 0\\ 0 & 0 & 1 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 \end{array}\right).$$
Because $P_{\pi}$ is obtained from the identity by swapping columns, its determinant will be either $1$ or $-1$; it is $1$ if you performed an even number of column exchanges/swaps, and $-1$ if you performed an odd number of column/swaps exchanges.
How does the parity of the number of column exchanges/swaps relate to $\pi$?