Notations.
Let $E$ be a subspace of dimension $k$ of $\mathbb R^n$, let $(e_1,\ldots,e_k)$ be an orthonormal basis of $E$.
Let $(X_1,\ldots,X_{k-1})$ be an orthonormal family of $E$. We can write the $X_i$ in terms of the $e_j$:
$$\forall i\in\{1,\ldots,k-1\},\quad X_i=\sum_{j=1}^k x_{i,j}e_j.$$
Since $\dim E=k$, there is a unique vector $Y\in E$ such that $(X_1,\ldots,X_{k-1},Y)$ is orthonormal and $\det(X_1,\ldots,X_k)>0$.
We can then define
$$f_1,\ldots,f_k\colon K\to \mathbb R$$
where $K$ is the compact of $\mathbb R^{k(k-1)}$ where $(x_{1,1},\ldots,x_{k-1,k})$ correspond to an orthonormal family $(X_1,\ldots,X_{k-1})$, such that $f_i$ is the $i$-th coordinates of $Y$ in the basis $(e_1,\ldots,e_k)$, i.e.
$$Y=\sum_{i=1}^k f_i(x_{1,1},\ldots,x_{k-1,k})e_i.$$
The question.
Let $i\in\{1,\ldots,k\}$.
Is $f_i$ Lipschitz continuous on $K$?
This means: does there exist $L$ such that
$$\forall x,y\in K,\quad \vert f_i(x)-f_i(y)\vert\leqslant L\Vert x-y\Vert?$$
What we know.
We know that the $f_i$ are continuous, and since $K$ is compact, Heine's theorem tells us that the $f_i$ are uniformly continuous. This is why I have great hopes that they will be Lipschtiz continuous, but I don't know how to prove it, and it seems really complicated to do a direct computation of the $f_i$.
Best Answer
Consider $k\times k$ matrix $X=\|x_{ij}\|$. By othonormality of the family $(X_i)$, we have $XX^t=I$, where $X^t$ is the transpose of the matrix $X$. Thus $(\det X)^2=\det X\cdot \det X^t=\det I=1$. Since $\det X$ is a positive real number, it equals $1$. Thus $X=(X^t)^t=(X^{-1})^t=C$ (see, for instance, Wikipedia), where $C=\|C_{ij}\|$ is the cofactor matrix of $X$. It follows that for each $l$, $f_l=C_{k,l}$, so $f_l$ is a polynomial function depending of variables $x_{i,j}$, $1\le i\le k-1$, $1\le j\le k$ defined on a compact subset $K$ of $\Bbb R^{k(k-1)}$, thus $f_l$ is Lipschitz.