The eigenvalues are exactly the same. The eigenvectors are the the coordinate vectors relative to $\mathcal{D}$ of the original eigenvectors.
Intuition. The matrix representation of $A$ relative to another basis gives the same linear transformation, just in a slightly different language; you are just doing a "change of coordinates"; but the action on the vectors (not on their names, but the actual vectors) stays the same. Since the action is the same, the eigenvalues and eigenvectors are the same, just "translated" into the new coordinates. Scalars (eigenvalues) don't need to be translated, so they stay the same.
Put another way: changing the basis and finding the representations relative to the new basis is like translating to a different language. If you take a novel or play written in English (the standard basis), and translate it into, say, Spanish (the different basis $\mathcal{D}$), the names of the characters may change (Henry may become Enrique, Joan of Arc may become Juana de Arco), but the characters will still perform the same actions in Spanish as they did in English. The plot doesn't change just because you translated, though the words used to described the plot did change.
Proof. To see that the eigenvalues are the same, note that $\lambda$ is an eigenvalue of $A$ if and only if $(A-\lambda I)\mathbf{x}=\mathbf{0}$ has a nontrivial solution. If $\mathcal{D}$ is a new basis for $V$, and we let $Q$ be the change-of-basis matrix from $\mathcal{D}$ to the standard basis (that is, the matrix whose columns are the vectors of $\mathcal{D}$), then the coordinate matrix of $A$ relative to the basis $\mathcal{D}$ is $Q^{-1}AQ$.
If $\mathbf{x}\neq \mathbf{0}$ is a solution to $(A-\lambda I)\mathbf{x}=\mathbf{0}$, then $Q^{-1}\mathbf{x}\neq \mathbf{0}$ (since $Q$ is invertible), and
$$(Q^{-1}AQ - \lambda I)(Q^{-1}\mathbf{x}) = Q^{-1}AQQ^{-1}\mathbf{x}-\lambda Q^{-1}\mathbf{x} = Q^{-1}A\mathbf{x} - Q^{-1}\lambda \mathbf{x} = Q^{-1}(A-\lambda I)\mathbf{x}=\mathbf{0},$$
so $\lambda$ is also an eigenvalue of $Q^{-1}AQ$.
Now repeat the argument with $B=Q^{-1}AQ$, but going from $\mathcal{D}$ to the original basis to conclude that if $\lambda$ is an eigenvalue of $Q^{-1}AQ$, then it is also an eigenvalue of $A$.
So the eigenvalues are identical.
And as we saw above, if $\mathbf{x}$ is an eigenvector of $A$ corresponding to $\lambda$, then $Q^{-1}\mathbf{x}$ is an eigenvector of $Q^{-1}AQ$ corresponding to $\lambda$; but $Q^{-1}\mathbf{x}$ is just the coordinate vector of $\mathbf{x}$ with respect to the basis $\mathcal{D}$.
Example. Take
$$A = \left(\begin{array}{cr}
-3 & -8\\
4 & 9
\end{array}\right).$$
The determinant of the matrix is $5$ and the trace is $6$, so the eigenvalues add up to $6$ and multiply out to $5$. Hence the eigenvalues are $\lambda=5$ and $\lambda=1$. The eigenvectors corresponding to $1$ are all nonzero multiples of $(2,-1)$; the eigenvectors corresponding to $5$ are the nonzero multiples of $(1,-1)$.
Now pick a new basis, say $\mathcal{D}=[(1,1), (-1,1)]$. Relative to this basis, we have that the change-of-basis matrix from $\mathcal{D}$ to the standard basis is the matrix $Q$ whose columns are the vectors of $\mathcal{D}$, so:
$$Q = \left(\begin{array}{rr}
1 & -1\\
1 & 1\end{array}\right),\qquad Q^{-1} = \left(\begin{array}{rr}
\frac{1}{2} & \frac{1}{2}\\
-\frac{1}{2} & \frac{1}{2}
\end{array}\right),$$
so the coordinate matrix of (the linear transformation given by multiplication by) $A$ relative to $\mathcal{D}$ is $[A]_{\mathcal{D}}$, where
$$[A]_{\mathcal{D}} = Q^{-1}AQ = \left(\begin{array}{rr}1 & 0\\12 & 5\end{array}\right).$$
You can see that the eigenvalues are still $1$ and $5$.
As for eigenvectors, the coordinate matrices of the old eigenvectors relative to $\mathcal{D}$ are:
$$
Q^{-1}\left(\begin{array}{r}2\\-1\end{array}\right) = \left(\begin{array}{r}\frac{1}{2}\\ -\frac{3}{2}\end{array}\right),\qquad\text{and}\qquad
Q^{-1}\left(\begin{array}{r}1\\-1\end{array}\right) = \left(\begin{array}{r}0\\-1\end{array}\right)$$
and you can verify easily that the eigenvectors of $[A]_{\mathcal{D}}$ corresponding to $5$ are nonzero multiples of $(\frac{1}{2}, -\frac{3}{2})$; and the eigenvectors of $[A]_{\mathcal{D}}$ corresponding to $1$ are the nonzero multiples of $(0,-1)$. (Which are "really" the same as the original vectors, only described "in Spanish" [relative to $\mathcal{D}$] instead of "in English" [relative to the standard basis]).
Best Answer
If you read $T_1$ from the last row to the first, you get $T_2$. If $x$ is an eigenvector of $T_1$ and $y$ an eigenvector for $T_2$ for the same eigenvalue, you get $$ x_k = y_{n-k+1} $$