A linear map that sends vectors to orthonormal basis is bounded Let $\{v_i\}_{i=1}^{d}$ be the set of independent vectors such that  $span\{v_1, \ldots, v_d\}=\mathbb{R}^d.$ It is well-known there is a linear map $L:\mathbb{R}^{d} \to \mathbb{R}^{d}$ that sends $\{v_i\}_{i=1}^d$ to the orthonormal basis.
$$Question$$
Is there $C>0$ such that $\|L\|<C$ Where $\|.\|$ is the operator norm?
 A: Any linear map between finite-dimensional normed vector spaces is bounded. In the case of $L:\Bbb{R}^n\to\Bbb{R}^m$, this follows easily from the Heine-Borel theorem (which implies compactness of the unit spheres). The general case follows by equivalence of norms to reduce it to the case of $\Bbb{R}^n\to\Bbb{R}^m$.
Note that if you decide to use a concrete norm like $1$-norm or $2$-norm or max-norm (i.e $\infty$-norm) on $\Bbb{R}^n$ and $\Bbb{R}^m$, then you can avoid appeals to any abstract theorems and directly prove the bound by fixing a matrix representation (with respect to the standard bases) and getting a bound (using very simple things like triangle inequality and perhaps Cauchy-Schwarz) in terms of the matrix entries.

Orthonormality or lack thereof is a red-herring.
A: The previous answer is off course correct and give a nice solution to your question. However there are some problems in the question. What is THE orthonormal basis of $\mathbb{R}^d$ ? There are lots of orthonormal basis in $\mathbb{R}^d$, but I could assume you mean THE canonical basis of $\mathbb{R}^d$ which is also orthonormal for the classical dot product $x \cdot y  = \sum_{i=1}^d x_i y_i $ ? If it is the case, may be your linear map L associates to $v^i$ the vector $e^i$ defined such as $e^i_j = \delta_{ij}$ ? Choosing for instance the euclidian $\|\cdot\|_2$ norm for vector then $\|Lx\|^2_2 = \sum_{i=1}^d (Lx)_i^2$ but
$$
Lx = L \left( \sum_{k=1}^d  x_k v^k \right) = \sum_{k=1}^d x_k (L v^k) = \sum_{k=1}^d x_k e^k 
$$
But $\{e^k\}_{k=1}^d$ forms an orthogonal basis of $\mathbb{R}^d$ (THE canonical basis), thus
$ (Lx)_i = Lx \cdot e^i = \sum_{k=1}^d x_k e^k \cdot e^i = x_i $ since $e^i \cdot e^k = \delta_{ij} $
Therefore, $(Lx)^2 = \sum_{i=1}^d x_i^2$ and you can majorate trivially the operator norm by one.
