From the product of two permutation matrices raised to the same power, is it easy to find the power? Let $A$ and $B$ be two public permutation matrices.
If $r$ is a secret power of large number, can we easily find $r$ from $A^rB^r$?
The product of two permutation matrices is itself a permutation matrix. So there's nothing special about its "factors" being public.  You should consider a single public permutation matrix $A$ and finding $r$ from $A^r$.


  Edit: oops, this was based on a misinterpretation of the question.


And there's nothing particularly special about it being a matrix. If $A$ is a permutation matrix corresponding to permutation $\pi$, then $A^r$ is the permutation matrix corresponding to $\pi$ composed with itself $r$ times. Let's call this permutation $\pi^{(r)}$.

So the question boils down to to the "discrete log" problem in a permutation group: given $\pi$ and $\pi^{(r)}$, find $r$. This problem is easy. 


First, observe that if $\pi$ is a simple cycle on $n$ items then $\pi^{(r)} = \pi^{(r \bmod n)}$. Since $n$ must be small in this problem, you can easily just "brute force" to determine $r$ mod $n$.
Second, any permutation can be uniquely decomposed into disjoint cycles. The operation $\pi^{(r)}$ acts independently on all the cycles of $\pi$. So if $\pi$ is composed of cycles of length $k_1, k_2, \ldots$ then we can use the above trick on each cycle and learn $r \bmod k_1, r \bmod k_2$ and so on. Then you can use the CRT to solve for $r \bmod \textrm{lcm}(k_1,k_2,\ldots)$. And indeed, the order of $\pi$ will also be $L=\textrm{lcm}(k_1,k_2,\ldots)$ in the sense that $\pi^{(L)}$ is the identity permutation. 


So you can efficiently identify $r$ as uniquely as is possible (mod $L$).