Let's take a supersingular elliptic curve $E / \mathbb{F}_q$, where $q = p^2$ for some $p \equiv 1 (\text{mod } 4)$, and $p \geq 5$, let's say. Then $ \# E(\mathbb{F}_q) = (p - 1)^2$, which by Hasse's theorem implies that the $q$-power Frobenius endomorphism $\phi_q$ satisfies the characteristic equation:
$\phi_q^2 - 2p \cdot \phi_q + q = 0 = (\phi_q - p)^2$
in $\text{End}(E)$.
How does this not imply that $\phi_q$ is identically equal to the multiplication-by-p map $[p]$? It's surely not, or else $\phi_q$ would represent an integral element of the quaternion-order $\text{End}(E)$.
EDITED: Looks like I flipped a sign in the trace equation (thanks @reuns). Thus requiring now $p \equiv 1 (\text{mod } 4)$, re: @hutner's comment, at least we have no contradiction in this case, as indeed $E(\mathbb{F}_q)$ is a $p - 1$-group.
The corresponding fact for $p \equiv 3 (\text{mod } 4)$ is that $\phi_q = [-p]$ should act as the identity on $E(\mathbb{F}_q) \cong \left( \mathbb{Z} / (p + 1)\mathbb{Z} \right)^2$. But this in turn is equivalent to $[p + 1]$ being the zero map, so we're again ok.
Thus at least no contradiction follows from @Lord Shark the Unknown's claim! Thanks everyone.