Suppose I have an $n\times n$ Vandermonde matrix $V_1$ where the roots/points in $x \in \mathbb{R}^n$ are all distinct and matrix $V_1$ is of the form $$V_1= [1 \; x \; x^2 \ldots x^{n-1}]$$ where the powers are element-wise.
When can I guarantee that I can always find a transformation $U$ such that $$V_2=V_1U$$ with $V_2$ another Vandermonde matrix?
Simple case: since $V_1$ is invertible, such transformation is given by $U= V_1^{-1}V_2$.
Suppose I now introduce a new root $x_{n+1}$ and thus a new row in the Vandermonde matrix $V_1$ which has now dimension $n+1 \times n$. Can I still find a transformation $U$ such that $V_2=V_1U?$ If $x_{n+1}= x_i$ for some $i \in 1,\ldots, n$ then the $U$ found above is still a good one. But what for a different root?
The answer is not necessarily. If $V_1$ is of the form $V_1 = [1\ \ x\ \ \cdots \ \ x^{n-1}]$ where $x \in \Bbb R^m$ has distinct entries and and $m \geq n$, then the columns of $V_1$ are linearly independent and thus span an $n$-dimensional subspace of $\Bbb R^n$. In general, a matrix $M$ such that $V_2 = V_1 M$ exists if and only if the column space of $V_2$ is a subspace of the column space of $V_1$.
As a simple example, there exists no $2 \times 2$ matrix $M$ such that $V_2 = V_1 M$ if $$ V_1 = \pmatrix{1&0\\1&1\\1&2}, \quad V_2 = \pmatrix{1&-1\\1&0\\1&2}. $$