Let $U$ and $W$ both be vector spaces over field $\mathbb{F}$, with $B_{U} = \big\{u_{1}, \ldots, u_{m}\big\}$ a basis for $U$ and $B_{W} = \big\{w_{1}, \ldots, w_{n}\big\}$ a basis for $W$. Then, the set $B_{U\otimes W} = \big\{u_{i} \otimes w_{j} | 1 \leq i \leq m \,, 1 \leq j \leq n \big\}$ is a basis for the tensor product space $U \otimes W$.
Thus, two different bases for $V \otimes V$ can be obtained using two different bases for $V$. Let $B$ be the basis given in the problem description, with $b_{1} = (1,0,0)$, $b_{2} = (0,1,0)$, and $b_{3} = (0,0,1)$. Let a second basis for $V$ be the set $B' = \big\{b_{1}', b_{2}', b_{3}' \big\}$ where $b_{1}' = \tfrac{1}{\sqrt{2}}(1,0,1)$, $b_{2}' = \tfrac{1}{\sqrt{2}}(1,0,-1)$, and $b_{3}' = (0,1,0)$.
With these two bases for $V$, it is possible to form the two following bases for $V \otimes V$:
(1)
\begin{align} \mathcal{B}_{B,B} = \big\{& b_{1} \otimes b_{1} \,,\, b_{1} \otimes b_{2} \,,\, b_{1} \otimes b_{3} \,, \\ & b_{2} \otimes b_{1} \,,\, b_{2} \otimes b_{2} \,,\, b_{2} \otimes b_{3} \,, \\ & b_{3} \otimes b_{1} \,,\, b_{3} \otimes b_{2} \,,\, b_{3} \otimes b_{3} \big\} \\ \\ \text{and} \\ \\ \mathcal{B}_{B,B'} = \big\{& b_{1} \otimes b'_{1} \,,\, b_{1} \otimes b'_{2} \,,\, b_{1} \otimes b'_{3} \,, \\ & b_{2} \otimes b'_{1} \,,\, b_{2} \otimes b'_{2} \,,\, b_{2} \otimes b'_{3} \,, \\ & b_{3} \otimes b'_{1} \,,\, b_{3} \otimes b'_{2} \,,\, b_{3} \otimes b'_{3} \big\} \end{align}
Noting that
(2)
\begin{align} b'_{1} & = \tfrac{1}{\sqrt{2}} \big( b_{1} + b_{3} \big) \,, \\ b'_{2} & = \tfrac{1}{\sqrt{2}} \big( b_{1} - b_{3} \big) \,, \quad \textrm{and} \\ b'_{3} & = b_{2} \end{align}
it is possible to write the $\mathcal{B}_{B,B'}$ basis exclusively in terms of the $b_{i}$ basis vectors:
(3)
\begin{align} b_{1} \otimes b'_{1} & = \tfrac{1}{\sqrt{2}} \big( b_{1} \otimes b_{1} + b_{1} \otimes b_{3} \big) \,, \\ b_{1} \otimes b'_{2} & = \tfrac{1}{\sqrt{2}} \big( b_{1} \otimes b_{1} - b_{1} \otimes b_{3} \big) \,, \\ b_{1} \otimes b'_{3} & = b_{1} \otimes b_{2} \,, \\ b_{2} \otimes b'_{1} & = \tfrac{1}{\sqrt{2}} \big( b_{2} \otimes b_{1} + b_{2} \otimes b_{3} \big) \,, \\ b_{2} \otimes b'_{2} & = \tfrac{1}{\sqrt{2}} \big( b_{2} \otimes b_{1} - b_{2} \otimes b_{3} \big) \,, \\ b_{2} \otimes b'_{3} & = b_{2} \otimes b_{2} \,, \\ b_{3} \otimes b'_{1} & = \tfrac{1}{\sqrt{2}} \big( b_{3} \otimes b_{1} + b_{3} \otimes b_{3} \big) \,, \\ b_{3} \otimes b'_{2} & = \tfrac{1}{\sqrt{2}} \big( b_{3} \otimes b_{1} - b_{3} \otimes b_{3} \big) \,, \\ b_{3} \otimes b'_{3} & = b_{3} \otimes b_{2} \,. \end{align}