Let $u, v \in V\doteq \mathbb{F}_2^{2 \times 2 \times 2}= \mathbb{F}_2 \otimes \mathbb{F}_2 \otimes \mathbb{F}_2$ be given by
$$u = e_1 \otimes e_1 \otimes e_1 + e_2 \otimes e_2 \otimes e_1 + e_1 \otimes e_2 \otimes e_2$$
and
$$v = (e_1+e_2) \otimes (e_1+e_2) \otimes (e_1+e_2) + e_1 \otimes e_2 \otimes (e_1+e_2) + e_2 \otimes e_1 \otimes e_2,$$
where $e_1, e_2$ are the usual standard basis vectors and $\otimes$ is the usual tensor product.
How can I construct an element $A \in {\rm GL}(V)$ such that $Au = v$? What is the explicit element $A$ that transforms $u$ to $v$? Thank you.
First of all, simplify $v$ using multilinearity of $\otimes$ together with the relations $e_1+e_1=e_2+e_2=0$ (since you are working over the field $\Bbb F_2$). In what follows, terms with the same colors will cancel: $$\begin{align} v &= e_1\otimes e_1\otimes e_1+e_1\otimes e_1\otimes e_2 + \color{blue}{e_1\otimes e_2\otimes e_1} + e_2\otimes e_1\otimes e_1 \\ &\quad+\color{green}{e_1\otimes e_2\otimes e_2} + \color{red}{e_2\otimes e_1\otimes e_2}+e_2\otimes e_2\otimes e_1 +e_2\otimes e_2\otimes e_2 \\ &\quad + \color{blue}{e_1\otimes e_2\otimes e_1}+\color{darkgreen}{e_1\otimes e_2\otimes e_2} + \color{red}{e_2\otimes e_1\otimes e_2} \end{align}$$Further grouping terms, we get $$v=e_1\otimes e_1\otimes e_1+e_1\otimes e_1\otimes e_2 + e_2\otimes e_2\otimes e_1+e_2\otimes e_2\otimes e_2 + e_2\otimes e_1\otimes e_1$$Now, if $e_{ijk} = e_i\otimes e_j\otimes e_k$, consider the ordered basis of $(\Bbb F_2)^{\otimes 3}$ given by $$\mathcal{B} = \{e_{111}, e_{112}, e_{121}, e_{122}, e_{211}, e_{212}, e_{221}, e_{222}\}$$Then $$[u]_{\mathcal{B}} = \begin{pmatrix} 1 \\ 0 \\ 0 \\ 1 \\ 0 \\ 0 \\1 \\ 0\end{pmatrix}\quad \mbox{and}\quad [v]_{\mathcal{B}} = \begin{pmatrix} 1 \\1 \\ 0 \\ 0 \\1 \\ 0 \\ 1 \\ 1\end{pmatrix}.$$Given any $8\times 8$ matrix $A \in {\rm Mat}(8, \Bbb F_2)$ which sends the first column vector into the second one, there is an unique $F \in {\rm Lin}((\Bbb F_2)^{\otimes 3})$ such that $[F]_{\mathcal{B}} = A$. The map you're looking for won't be unique because the matrix $A$ will also not be. Then you only have to pick a matrix which is invertible over $\Bbb F_2$. Here's one concrete matrix doing this $$A = \begin{pmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \end{pmatrix}$$Identifying blocks of orders $2$, $2$ and $3$ and multiplying their determinants, it is easy to see that $\det A = 1$, so the map $F$ corresponding to $A$ is actually in ${\rm GL}((\Bbb F_2)^{\otimes 3})$. Here's how I came up with this matrix. I searched in lower dimension for non-singular matrices taking $$\begin{pmatrix} 1 \\ 0 \end{pmatrix}\mapsto \begin{pmatrix} 1 \\ 1 \end{pmatrix}, \quad \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix} \mapsto \begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}\quad\mbox{and}\quad \begin{pmatrix} 0 \\ 1 \\ 0\end{pmatrix} \mapsto \begin{pmatrix} 0 \\ 1 \\ 1\end{pmatrix}.$$The matrix $A$ above is the direct sum of said smaller matrices. Concretely, the map $F$ is the unique linear endomorphism of $(\Bbb F_2)^{\otimes 3}$ satisfying $$\begin{align} F(e_1\otimes e_1\otimes e_1) &= e_1\otimes e_1 \otimes(e_1+e_2) \\ F(e_1\otimes e_1\otimes e_2) &= e_1\otimes e_1\otimes e_2 \\ F(e_1\otimes e_2\otimes e_1) &= e_1\otimes e_2\otimes e_1 \\ F(e_1\otimes e_2\otimes e_2) &= e_2\otimes e_1 \otimes e_1 \\ F(e_2\otimes e_1\otimes e_1) &= e_1\otimes e_2\otimes e_2 \\ F(e_2\otimes e_1\otimes e_2) &= e_2\otimes e_1\otimes e_2 \\ F(e_2\otimes e_2\otimes e_1) &= e_2\otimes e_2\otimes (e_1+e_2) \\ F(e_2\otimes e_2\otimes e_2) &= e_2\otimes e_2\otimes e_2.\end{align}$$