Let $F$ be a field, and $A$ a rank-one matrix in $M_n(F)$. Let $A$ is similar to $E_{12}$ where $E_{12}$ denotes the matrix with 1 in the row $1$th and the column $2$th, and zero elsewhere.
Now, we want find a especial basis and the matrix of change of basis that $A$ changes to $E_{12}$.
On
Assume that you have two sets of linearly independent vectors $\{b_k\}$ and $\{c_k\}$, which span the same vector space. Define the matrices $B=\big[b_1 b_2\ldots b_n\big]$ and $C=\big[c_1 c_2\ldots c_n\big]$ whose columns are these basis vectors.
A matrix $M$ which converts between these two bases must satisfy $n$ vector equations $$\eqalign{ Mb_k &= c_k \cr MB &= C \implies M &= CB^{+} \cr }$$ where $B^{+}$ is the pseudoinverse of $B$; if $B$ is square, then it's the standard inverse.
A rank-1 matrix is simply the outer product of two vectors, so $A=xy^T$.
These vectors can be obtained from the SVD of $A$.
You want to calculate $M$ having specified the first two vectors in each basis, i.e.
$$\eqalign{
b_1&=x &\,\,\,b_2=y \cr
c_1&=e_1 &\,\,\,c_2=e_2 \cr
}$$
where $\{e_k\}$ are the columns of the identity matrix
and $\,\,e_1e_2^T=E_{12}$
The exact expression for $M$ will depend on the remaining columns of $B$ and $C$.
So if I understand the question correctly:$$E_{12}=\begin{bmatrix}0&1\\0&0\end{bmatrix}$$$$A=\begin{bmatrix}0&a\\0&0\end{bmatrix}$$with $a\in F$. If this is the case, then we're able to use a property of the field:
For each $x\in F$ there exists an inverse element $x^{-1}\in F$ such that $xx^{-1}=1$. Then to change $A$ to $E_{12}$ we would like to multiply by $a^{-1}$. Can you take it from here?