Knot Theory: Mutations Show that if we have three tangles as in Figure 2.33a, we can mutate several times in order to permute the tangles. 
Note that we can then permute n tangles in a row. This is from Colin Adams; The Knot book exercise 2.25.

 A: Denote a mutated $T_{i}$ by $T^{i}$ for convenience. Both $T_{2}$ and $T_{3}$ intersect a small region around themselves exactly 4 times. Consider then the region around both of them together. It also only intersects the region four times. Mutate this region. It is then $T^{3}$ and $T^{2}$. So our order of tangles is $T_{1},T^{3},T^{2}$. We can then mutate only $T^{3}$ to get the order $T_{1},T_{3},T^{2}$. Then we simply mutate $T^{2}$ to get the desired order $T_{1},T_{2},T_{3}$. I think this is correct based simply on the definition of mutation from wikipedia. I have not interacted with mutations before, very interesting. 
A: To complement Jeremy's answer, you can also do the following. I'll use the same notation as it is convenient, keeping in mind that we will always mean left-to-right reflection when we say mutation.
First note that $\overline{T_1T_2T_3}$, the closure of the composition of the tangles $T_1,T_2$ and $T_3$, is equal to $\overline{T_2T_3T_1}$ as cyclic permutations respect closure of tangles. We may now mutate the subtangle $T_2T_3T_1$ to get the knot $\overline{T^1T^3T^2}$ and now if we mutate each component subtangle individually we get the knot $\overline{T_1T_3T_2}$ as required.
This relies on the following easy to prove result that if $\mu$ is the left-to-right reflection mutation, then if $T_i$ and $T_j$ are tangles, $\mu(T_iT_j)=\mu(T_j)\mu(T_i)$. Algebraically this says that $\mu$ is an anti-automorphism on the monoid of tangles (with product being horizontal composition of tangles and identity being two vertical straight line segments).
