For a function $f(x)=[f_1(x),f_2(x),\cdots,f_m(x)]^T:\mathbb{R}^n\to\mathbb{R}^m$ where $f_i(x):\mathbb{R}^n\to \mathbb{R}$ is a real-value function, $f(x)$ is Lipschitz, i.e., there exists a positive real constant $L$ such that $\|f(x)-f(y)\|\leq L\|x-y\|$.
Then, the question is whether there exist the following inequations $$\|f^{\mu}(x)-f^{\mu}(y) \|\leq L^{\mu}\|x-y\|^{\mu}, \forall \mu>1,$$ and $$\|f^{\mu}(x)-f^{\mu}(y)\| \leq L^{\mu}\|x-y\|^{\mu}, \forall \mu<1$$ where $f^{\mu}(x)=[f_1^\mu(x),f_2^\mu(x),\cdots,f_m^\mu(x)]^T$.
Thanks very much.
Clearly we need $f\ge0$ if $\mu$ is not an integer, which I will assume.
Simple counterexample in dimension 1. let $f:\mathbb R\to\mathbb R$, $f(x)=x$, $y=1$, and $\mu=2$. Then the wanted inequality is $$ |x^2 - 1| \stackrel{?}{\le} C(x-1)^2$$ Lets put $x=1+\epsilon$, for $\epsilon>0$. Then this reads $$ \epsilon^2 + 2\epsilon\stackrel{?}{\le} C\epsilon^2 $$ which is easily seen to be wrong for any $C>0$, once $\epsilon\ll1$.
In fact, the inequality implies that $f$ is everywhere differentiable with derivative $0$, so the only such $f$ are (locally) constant.
This is true, at least up to a constant. Recall that $x^\alpha$ is $\alpha$-Hölder with constant 1. So for each coordinate, we have $$ |f_i^\mu(x) - f_i^\mu(y)| \le |f_i(x) - f_i(y)|^{\mu} \le L^\mu |x-y|^\mu$$ which implies the result with a potentially different constant depending on the precise choice of norm $\|\cdot\|$.