The condition number of a scaled Vandermonde matrix

Let $V(x_1,..,x_n)$ be the Vandermonde matrix induced by $x_1,..,x_n$, and let $\tilde{V} := V(\frac{x_1}{h},...,\frac{x_n}{h})$. My intuition says that the condition number should be invariant under such scaling of the nodes at least for some special cases of node configurations. My question is then:

1. Are there constraints on the nodes $x_1,..,x_n$ (i.e symmetric, positive, equally spaced or others) for which ${\cal K}(\tilde{V})= {\cal K}(V)$, where ${\cal K}(V)$ is the condition number for a matrix $V$?
2. Are there known lower and upper bounds which relate the singular values of $V$ and $\tilde{V}$ to one another?
• For any configuration of nodes, as $h\to \infty$, $\tilde{V}$ tends to a singular matrix while remaining bounded in norm, so $\mathcal K(\tilde{V})\to \infty$. – Mike Jury Nov 12 '15 at 15:44
• This comment pretty much sums it up. If you can repost it as an answer, I can accept and close. Thanks. – gil Nov 18 '15 at 12:45

For any configuration of nodes, as $h\to \infty$, $\tilde{V}$ tends to a singular matrix while remaining bounded in norm, so $\mathcal K(\tilde{V})\to \infty$.
$$\tilde{V}D = V,$$
where $D = diag(1,h,h^2\ldots,h^{n-1})$. But the 2-norm condition number of a matrix is not invariant under scaling of the columns. Thus, $\mathcal{K}(\tilde{V})\neq\mathcal{K}(V)$.