I'm not quite able to show this. Some playing around says that there doesn't exist two bases such that the matrices are the same (correct me if I'm wrong). I am just trying to consider where the standard basis maps to. D maps things off diagonal 'one step to the right', whereas M maps things 'one step to the left'. However I'm not sure how to show the matrices can't be the same for any basis.
Since their dimensions are the same, $V_1,V_2$ are isomorphic.
Suppose there exists such $B_1, B_2$ such that the matrix representation for $D$ and $M$ is the same. $$B_1=[p_1,\ldots,p_n]$$ $$B_2=[q_1,\ldots,q_n]$$ where $p_i, q_i$ are polynomials of degree less than 1.
$B_2$ is a basis for $V_2$, so there exists $q_{k_j}$ such that $q_{k_j}$ contains a $a_jx^{n-2}$ term, for $j=1,\ldots,m$.
Then for an arbitrary vector $$v= [p_1,\ldots,p_n] [x_1,\ldots,x_n]^T= [q_1,\ldots,q_n] [y_1,\ldots,y_n]^T $$ we have $$\sum_{i=1}^n x_iD(p_i) = \sum_{i=1}^n y_ix(q_i)$$
The latter is modulo $x^n$ form.
Since $deg p_i < n$, $deg D(p_i) < n-1$. So we can deduce that for any vector $v$, $\sum_{j=1}^m y_ja_j$ is zero, otherwise from the right side we will have $deg M(v)=n-1$. This contradicts our assumption $v$ is arbitrary.