Extra. Let's start again with the above parametrization, written as: $$ x - s_x = v_x \cdot t + a_x \cdot \frac{1}{2} t^2 \\ y - s_y = v_y \cdot t + a_y \cdot \frac{1}{2} t^2 $$ Now put: $$ \xi = t \quad ; \quad \eta = \frac{1}{2} t^2 \quad \Longrightarrow \quad \eta = \frac{1}{2} \xi^2 $$ And: $$ x - s_x = v_x \, \xi + a_x \, \eta \\ y - s_y = v_y \, \xi + a_y \, \eta $$ Two equations with two unknowns: $$ \xi = \frac{+ a_y (x-s_x) - a_x (y-s_y)}{v_x a_y - v_y a_x} \qquad ; \qquad \eta = \frac{- v_y (x-s_x) + v_x (y-s_y)}{v_x a_y - v_y a_x} $$ Substitute into the "normed parabola" $\;\eta = \frac{1}{2} \xi^2\;$ or $\;\frac{1}{2} \xi^2 - \eta = 0$ : $$ \frac{1}{2} \left[ \frac{a_y (x-s_x) - a_x (y-s_y)}{a_y v_x - a_x v_y} \right]^2 + \left[ \frac{v_y (x-s_x) - v_x (y-s_y)}{a_y v_x - a_x v_y} \right] = 0 $$ Which is the equation of an arbitrary parabola in $x$ and $y$ alone.
Update. But, if I read the question well, it is required that the transformation taking the parabola $p_1$ into $p_2$ is represented by a scalar times an orthogonal matrix. This is most easily achieved by a parameter shift in: $$ x(t) = \frac{1}{2} a_x.t^2 + v_x.t + s_x \\ y(t) = \frac{1}{2} a_y.t^2 + v_y.t + s_y $$ As follows: $$ \overline{x}(t) = \frac{1}{2} a_x(t-\tau)^2 + v_x(t-\tau) + s_x \\ \overline{y}(t) = \frac{1}{2} a_y(t-\tau)^2 + v_y(t-\tau) + s_y $$ $$ \overline{x}(t) = \frac{1}{2} a_x.t^2+\left(v_x-a_x.\tau\right)\,t +\left(\frac{1}{2}a_x.\tau^2-v_x.\tau+s_x\right) \\ \overline{y}(t) = \frac{1}{2} a_y.t^2+\left(v_y-a_y.\tau\right)\,t +\left(\frac{1}{2}a_y.\tau^2-v_y.\tau+s_y\right) $$ $$ \overline{x}(t) = \frac{1}{2} a_x.t^2 + \overline{v}_x.t + \overline{s}_x \\ \overline{y}(t) = \frac{1}{2} a_y.t^2 + \overline{v}_y.t + \overline{s}_y $$ Now it's always possible to choose $\tau$ in such a way that $(\overline{v}_x,\overline{v}_y)$ is perpendicular to $(a_x,a_y)$ : $$ a_x.\left(v_x-a_x.\tau\right) + a_y \left(v_y-a_y.\tau\right) = 0 \quad \Longrightarrow \quad \tau = \frac{v_x a_x + v_y a_y}{a^2_x + a^2_y} $$ This means that the local coordinate systems $(\vec{v},\vec{a})$ of our parabolas have become orthogonal. And a transformation that takes an orthogonal coordinate system into another orthogonal coordinate system is itself orthogonal (apart from some scalar eventually).