Let $q \equiv 1 \pmod{4}$ be prime, let $E/\mathbb{F}_q : y^2 = x^3 + ax$ be an elliptic curve and let $i$ be a number satisfying $i^2 = -1$. Then, the map $\phi : E \to E$ such that $\phi(x,y) = (-x, iy)$ is an endomorphism defined over $\mathbb{F}_q$.
I have seen in many places Example 4 of GLV paper and Pairings for Beginners that if you take a point $P \in E(\mathbb{F}_q)$ of primer order $r$, then $\phi(Q) = [\lambda]Q$ (so, $\phi$ acts as a multiplication-by-$m$ map on $E$) for all $Q \in \langle P \rangle$, with $\lambda$ an integer satisfying $\lambda^2 = -1 \pmod{r}$. Why is this true? I know that $\phi \circ \phi = [-1]$, but I do not see why it acts as a multiplication-by-$m$ map.
You already noted $\phi \circ \phi = [-1]$. In other words $\phi^2 = [-1]$, where the identity lives in the (noncommutative) ring of endomorphisms of $E$. Now take $\lambda \in \mathbb{F}_q$ such that $\lambda^2= -1$. Then also $[\lambda]^2 = [\lambda^2] = [-1]$.
In particular $\phi^2 = [\lambda]^2$, which we rewrite as: $$(\phi - [\lambda])(\phi + [\lambda]) = \phi^2 - [\lambda]^2 = 0$$ Now it is a (nontrivial) fact that the ring of endomorphisms of $E$ has no zero divisors. That implies that either $\phi = [\lambda]$ or $\phi = -[\lambda] = [-\lambda]$. In both cases, $\phi$ is multiplication by something that squares to 1.
(Note that this is really just repeating the general proof that in a ring $R$ without zero divisors, any two square roots of $a \in R$ agree up to sign.)