Here's a possible strategy:
Note that the dimension of $\operatorname{im}(T)$ equals the rank of $A$ and hence the rank of $D$ since conjugate matices have same rank. What is the rank of $D$? Hence, what is the dimension of $\operatorname{im}(T)$? Given the eigenvectors corresponding to the distinct non-zero eigenvalues, what are obvious independent elements of $\operatorname{im}(T)$?
Further, how are the dimensions of $\operatorname{im}(T)$ and $\operatorname{ker}(T)$ related? Hence, given that $0$ is an eigenvalue, what is $\operatorname{ker}(T)$?
You can formulate change of basis rigorously as follows, given $S$ is a linear transformation. Let $[S]_{\beta}^{\beta '}$ be the matrix corresponding to $S$ with respect to basis $\beta$ in the domain and $\beta'$ in the codomain. In our case, if $\beta =\{v_1,...,v_n\}$and $\beta'=\{w_1,...,w_n\}$, then we can write $S(v_j)=\sum_{i}a_{ij}w_i$. Then $[S]_{\beta}^{\beta'}=\begin{bmatrix}a_{11} & a_{12}&\dots&a_{1n} \\a_{21}&a_{22}&\dots&a_{2n}\\\vdots&\vdots&\ddots&\vdots\\a_{n1}&a_{n2}&\dots&a_{nn} \ \end{bmatrix}$. Now matrix multiplication is compatible with linear transformation if the bases are in the "right place". For example if $S$ and $T$ are linear transformation (from $\mathbb R^n$ to itself), then given $\beta,\beta',\beta''$ are bases of $\mathbb R^n$. Then $[ST]_{\color{blue}{\beta}}^{\color{green}{\beta ''}}=[S]_{\color{red}{\beta'}}^{\color{green}{\beta''}}[T]_{\color{blue}{\beta}}^{\color{red}{\beta '}}$
Now back to your question. Let's say you have a linear transformation $A$, and you know its matrix corresponding to a certain basis $\beta$, i.e. given $[A]_{\beta}^{\beta}$. You seek to find the matrix of $A$ corresponding to another basis $\beta'$, i.e. $[A]_{\beta'}^{\beta'}$. Then by the identity above, let $I$ be the identity map, we should have $[A]_{\beta'}^{\beta'}=[I]^{\beta'}_{\beta}[A]_{\beta}^{\beta}[I]_{\beta'}^{\beta}$. Specifically in your case, $\beta'=\{(1,2),(2,0)\}$, $[I]_{\beta'}^{\beta}$ is just asking how to write $a\begin{bmatrix}1\\2\end{bmatrix}+b\begin{bmatrix}2\\0\end{bmatrix}$ in terms of something like $f(a,b)\begin{bmatrix}1\\0\end{bmatrix}+g(a,b)\begin{bmatrix}0\\1\end{bmatrix}$. Then,
$$[I]_{\beta'}^{\beta}:\begin{bmatrix}a\\b\end{bmatrix}\mapsto\begin{bmatrix}f(a,b)\\g(a,b)\end{bmatrix}=\begin{bmatrix}a+2b\\2a\end{bmatrix}=\begin{bmatrix}1&2\\2&0\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix}$$,
i.e. $[I]_{\beta'}^{\beta}=\begin{bmatrix}1&2\\2&0\end{bmatrix}$
And then $[I]_{\beta}^{\beta'}=([I]_{\beta'}^{\beta})^{-1}=\begin{bmatrix}1&2\\2&0\end{bmatrix}^{-1}$, because their product should give you $\begin{bmatrix} 1&0\\0&1 \end{bmatrix}$.
But when you do the reversed direction, you need to find write $\begin{bmatrix}1\\0\end{bmatrix}$and $\begin{bmatrix}0\\1\end{bmatrix}$ in terms of $\begin{bmatrix}1\\2\end{bmatrix}$ and $\begin{bmatrix}2\\1\end{bmatrix}$, so it should be the reversed process. Again we start by writing any vector as $a\begin{bmatrix}1\\0\end{bmatrix}+b \begin{bmatrix}0\\1\end{bmatrix}$, the matrix should send this $(a,b)$ to the coefficients with respect to the other basis' vectors, i.e. $f'(a,b)\begin{bmatrix}1\\2\end{bmatrix}+g'(a,b)\begin{bmatrix}2\\0\end{bmatrix}$, which you can find by solving the linear equations or finding matrix's inverse.
In short, the "easy to remember method" for change of basis matrix $[I]_{\beta}^{\beta'}$ from $\beta=\{v_1,...,v_n\}$ to $\beta'=\{w_1,...,w_n\}$ is that the $\text{j}^{\text{th}}$ column is the column of unique coefficients to write $v_j=a_{1j}w_1+...+a_{nj}w_n$, i.e. $\begin{bmatrix} a_{1j}\\a_{2j}\\ \vdots\\a_{nj}\end{bmatrix}$
Best Answer
You get no information whatsoever about non-zero eigenvalues/eigenvectors from $[T]_{B,C}$ unless you know $B$ and $C$. Of course, knowing the matrix for an operator with respect to known bases, does allow you to reconstruct the operator, and hence information such as eigenvectors/eigenvalues. But, if you just have a matrix, with respect to two unknown bases, you have basically no information.
You do get a bit of information about eigenvalues/eigenvectors of $0$. Recall that the eigenvectors corresponding to $0$ are the kernel of $T$. The nullspace of the matrix $[T]_{B, C}$ is always the image of the kernel of $T$ mapped under the coordinate vector map with respect to $B$. That is, $$p \in \operatorname{ker} T \iff [p]_B \in \operatorname{null} [T]_{B, C}.$$ As the coordinate map is injective, this makes the two spaces isomorphic, and hence of the same dimension.
This map therefore takes eigenvectors of $T$ to eigenvectors of $[T]_{B, C}$, each corresponding to $0$. Note that this is the usual correspondence we get when considering eigenvectors of $T$ and eigenvectors of $[T]_{B, B}$ (for more general eigenvalues). In other words, the situation is largely unchanged for the eigenvalue $0$.
However, it's also worth noting that generalised eigenvectors corresponding to $0$ are free game. While the geometric multiplicity (the dimension of the eigenspace) is fixed, the algebraic multiplicity (the dimension of the generalised eigenspace, a.k.a. the exponent of the factor $\lambda$ in the characteristic polynomial) can definitely change. These extra dimensions can become new non-zero eigenvalues, be absorbed by other non-zero eigenvalues, or still contribute to the $0$ eigenvalue (possibly changing the structure of the Jordan Blocks corresponding to $0$).
I've given no examples of this, but here's something that may help. Pick your favourite invertible map $T : V \to V$, where $V$ is finite-dimensional. Pick your favourite basis $B = (b_1, b_2, \ldots, b_n)$ of $V$. Then $C = (Tb_1, Tb_2, \ldots, Tb_n)$ is also a basis! Further, using these bases, it's straightforward to show that $$[T]_{B, C} = I_{n \times n},$$ i.e. the identity matrix. So, any invertible map, with any array of eigenvalues and eigenvectors, can become totally homogenised to the point of being the identity map. Any subtleties about the structure of the eigenspaces (e.g. diagonalisability) is totally gone, and now the whole space is one big, undifferentiated eigenspace corresponding to the single eigenvalue $1$.
In other words, if you care about eigenvalues/eigenvectors, definitely consider $[T]_{B, B}$ or $[T]_{C, C}$. I hope that helps!