Let $f_i$ ($i=1, \ldots, n$) be globally Lipschitz continuous functions which can be expressed as
$\left[\begin{array}{c} f_1(x_1, \ldots, x_n) \\ \vdots \\ f_n(x_1, \ldots, x_n)\end{array}\right] = \left[\begin{array}{ccc} a_{11} & \cdots & a_{1n} \\ \vdots & \ddots & \vdots \\ a_{n1} & \cdots & a_{nn}\end{array}\right] \left[\begin{array}{c} g(x_1) \\ \vdots \\ g(x_n)\end{array}\right].$
where $a_{ii} \neq 0$ for all $i$.
Can we say that the scalar function $g$ is globally Lipschitz continuous?
Yes.
First, the case where there the line $j=1$ has a sum $\sum_i a_{1i}\neq 0$. Then we study on the diagonal $x := x_1 = x_2 = \dots = x_n$, and the answer is fast. $$g(x) = \frac {f_1(x,x,\dots,x)} {\sum_i a_{1i}}$$
Else, we study on the pseudo-diagonal $x := x_2 = \dots = x_n$ with $x_1 := 0$, which gives : $$g(x) = \frac {f_1(0,x,\dots,x) - a_{11}g(0)} {\sum_{i\ge 2} a_{1i}}$$ Where the denominator is not $0$ because it is equal to $-a_{11}$ because the line 1 has null sum.
In both case, we have easy proofs that $g$ has almost any regularity that the $f_i$ have.