Finding a conjugacy

Find a conjugacy of the form $\varphi(x)=ax+b$ from $f(x)=x^2-2$ to $g(x)=4x(1-x)$.

We must satisfy the relationship $$f\circ\varphi=\varphi\circ g$$ The left-hand side of the equation evaluates to $$f\circ\varphi=(ax+b)^2-2=a^2x^2+2abx+b^2-2$$ The right-hand side of the equation evaluates to $$\varphi\circ g=a(4x(1-x))+b=-4ax^2+4ax+b$$ For both sides of the equation to be equal, it must be the case that the coefficients are equal. So $a=-4$ and $b=2$.

Therefore, $$\varphi(x)=-4x+2$$

@RedCrayon Nice! Note also that
$$f\circ\varphi(x) = 16 x^2-16 x+2 = \varphi\circ g(x).$$

@mark, is this in general true?

@RedCrayon It's always true that $f\circ\varphi(x) = \varphi\circ g(x)$, by the definition of conjugacy. Note, though, that I fixed a typo in my last reply because I had said that $f\circ\varphi(x) = \varphi\circ f(x)$, which is rarely true.