Let $f(a,b,t)=tb+a(1-t)$. Then, we are looking for a value $t_2$ such that $f(a,b_1,t_1)=f(a,b_2,t_2)$ where $a$ represents the source value, $b_1$ represents the destination value, $t_1$ represents the interpolation factor, and $b_2$ and $t_2$ represent the new source and interplation factor respectively. If we substitute and rearrange: $$ \begin{align*} f(a,b_1,t_1)&=f(a,b_2,t_2);\\\ t_1b_1+a(1-t_1)&=t_2b_2+a(1-t_2);\\\ t_1(b_1-a)&=t_2(b_2-a);\\\ t_2&=\boxed{\frac{t_1(b_1-a)}{b_2-a}}, \end{align*} $$ which is the formula for the interpolation percentage such that the interpolated value stays the same.