We have the following the min cost network flow problem. Notice that the arcs $(1,2), (2,3), (4,2), (3,6), (5,6)$ give a feasible basis. We easily find a feasible solution:
$$ (x_{12}, x_{23}, x_{36}, x_{42}, x_{56}) = (-6,-1,-1,-3,-1) $$
where we have assumed $b_i < 0$ is supply . When trying to find dual variables, we obtain
\begin{align*} y_1-y_2 = c_{12} = 2 \\ y_2-y_3 = c_{23} = 3 \\ y_4-y_2 = c_{42} = 5 \\ y_3-y_6= c_{36} = 3 \\ y_5-y_6 = c_{56} = 7 \\ \end{align*}
My question is, in solving this system, can we assume $y_1$ or $y_5$ to be $0$ since we can for a root node from them?
Yup, that's fine, the notion of a root node is arbitrary, just grab the node that you want it to be the root node, shake it and with the help of gravity, it becomes the root node.
To see why we can do that in terms of the algorithm. What really matters is the difference, as only the difference of the dual variables/ potential values are used in computing the optimality conditons. We have $n$ potential values and $n-1$ arcs connecting them to form the tree, fixing a value would fully determined the rest. Fixing it differently would adjust two solutions by a constant.