The equation of motion for a free particle of mass $m$, moving along $x$-axis, is governed by the differential equation \begin{equation}m\frac{d^2x}{dt^2}=0.\tag{1}\end{equation}
A simple observation is that under a simple scaling of $x$ and $t$ i.e., $$(x,t)\to(\bar{x},\bar{t})=(\lambda x,\mu t)~~~~~~~~~~~~~~~~~~\lambda,\mu=\text{non-zero constants}\tag{2}$$ where becomes\begin{equation} m\frac{\mu^2}{\lambda}\frac{d^2\overline{x}}{d\overline{t}^2}=0\Rightarrow m\frac{d^2\overline{x}}{d\overline{t}^2}=0 \end{equation} which implies that the equation remains unchanged in form.
$\bullet$ What inference can be drawn about the solutions of (1) in the presence and in absence of a given initial condition?
$\bullet$ If we scaled only $t\to \mu t$, and left $x$ unchanged, it would have implied that if $x(t)$ is a solution of (1), so does $x(\mu t)$. I'm not sure how to interpret the meaning of the invariance when both $x$ and $t$ are scaled as in (2).
You can rewrite equation (1) as $$m\frac{dv}{dt}=0$$ wit the solution $v=const.$ So the first equation is Newton's first law. The scaling of $x$ and $t$ just means that the above statement remains the same if you change your units. Instead of using meters and seconds, one can use miles and hours, and a body at rest will be still at rest.