Please consider the link L6n1
Please note that such link is not the Borromean link L6a4.
I am trying to obtain the braid word for L6n1.
Using SnapPy with the following code
In[5]: L2=Link('L6n1')
In[6]: word2 = L2.braid_word(); word2
we obtain the output
Out[6]: [1, -2, 1, 2, -1, 2]
it is to say, the word is $$ {\sigma_{{1}}}{\sigma_{{2}}}^{-1}{\sigma_{{1}}}{\sigma_{{2}}}{\sigma_{{1}}}^{-1}{\sigma_{{2}}}$$
From other side in the paper https://arxiv.org/pdf/1104.5144.pdf
the braid word for L6n1 is given as
$$\sigma_{{1}}\sigma_{{2}}\sigma_{{1}}\sigma_{{2}}\sigma_{{1}}\sigma_{{2}} = (\sigma_{{1}}\sigma_{{2}})^ {3}$$
Then my question is: how to transform the first braid word in to the second braid word?
Those words don't represent the same element, so they cannot be transformed into each other. The reason is that they don't have the same exponent sum. The relation of the braid group $B_n$ are homogeneous, so any transformation using them keeps the exponent sum constant.
By homogeneous I mean that they have the same exponent sum at both sides, namely, if the generators are $\sigma_i$ ($i=1,\dots, n-1$) there are relations of the form $\sigma_i\sigma_j=\sigma_j\sigma_i$ (exponent sum equals 2 at both sides) and relations of the form $\sigma_i\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}$ (exponent sum equals 3 at both sides). For the case $n=3$, there are only relations of the second type.
As a side note, I checked that indeed your link is the same as the $B_3$ NUS Link in the paper, so it is probably a mistake of the writer (or the software, but I guess that's less likely).