= Solution
Because $\phi$ is a <separable isogeny>,
$$
\phi^*(O_{E_2})=\sum_{T\in E_1[\phi]}(T).
$$
The degree-zero divisor
$$
D=\phi^*(O_{E_2})-d(O_{E_1})
$$
corresponds under $E_1\simeq\operatorname{Pic}^0(E_1)$ to the sum of all elements of the finite abelian group $E_1[\phi]$. Pairing every $T$ with $-T$ shows that this sum is the sum of the elements in $E_1[\phi]\cap E_1[2]$. It vanishes when that intersection has one element and also when it has four elements, since the sum of the four elements of $(\mathbb Z/2\mathbb Z)^2$ is zero. The <principal divisor criterion on an elliptic curve> therefore gives a rational function $g$ with $\operatorname{div}(g)=D$.
The divisor $D$ is invariant under $[-1]$, so $[-1]^*g/g$ has zero divisor and is constant. Applying $[-1]$ twice shows that this constant has square one; hence $[-1]^*g=\pm g$.
For a short <Weierstrass equation of an elliptic curve>
$$
E:y^2=x^3+Ax+B,
$$
the multiplication-by-two isogeny has
$$
\operatorname{div}(y)=\sum_{T\in E[2]}(T)-4(O_E)=[2]^*(O_E)-4(O_E).
$$
Thus one may take $g=y$, and $[-1]^*y=-y$. For multiplication by three, the third <division polynomial of an elliptic curve>
$$
\psi_3(x)=3x^4+6Ax^2+12Bx-A^2
$$
vanishes simply at the eight nonzero points of $E[3]$ and has a pole of order eight at $O_E$. Hence $g=\psi_3(x)$ has divisor $[3]^*(O_E)-9(O_E)$ and satisfies $[-1]^*g=g$. Both signs occur.
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