If we consider a linear function $f$, and an arbitrary positive constant $\lambda$ and points $x,y$, then we can find the distance from $f(\lambda x)$ to $f(\lambda y)$ as
$\| f(\lambda x) - f(\lambda y)\| = \| \lambda f(x) - \lambda f(y)\| = \lambda \| f(x) - f(y)\|$
If we now relax the linearity condition, to at most linear growth, i.e. $\|f(x)\| \leq c \|x\|$ and possibly also to satisfy the Lipschitz condition. Can we then do something similar to obtain, something like the following?
$\| f(\lambda x) - f(\lambda y)\| \leq \lambda^{\alpha} \| f(x) - f(y)\|$
No, without linearity there is no reason for $\| f(\lambda x) - f(\lambda y)\|$ to be controlled by $\| f(x) - f(y)\|$. For example, let
$$ f(x) = \max(0, |x|-1),\qquad x\in\mathbb{R} $$ This is a Lipschitz continuous function which satisfies $|f(x)|\le |x|$. Yet, there is no constant $C$ such that $$ |f(1)-f(2)| \le C|f(0.1)-f(0.2)| $$ since the right hand side is zero while the left hand side is not.