I think the situation you describe is impossible: Let $\bar{\phi}$ be the conjugate endomorphism to $\phi$. From the positivity of the dimension and equation $d(\phi) d(\phi)^2 = 2d(\phi)$ we get $d(\phi) = 2$. Denote by $\langle \rho, \sigma \rangle$ the dimension of the intertwiner space between $\rho$ and $\sigma$. By Frobenius reciprocity and the irreducibility of $\phi$ we now have
$$ \langle \bar{\phi} \circ \phi, \phi \rangle = \langle \phi, \phi \circ \phi \rangle = \langle \phi, \phi \oplus \phi \rangle = 2. $$
Thus, $\bar{\phi} \circ \phi$ contains two copies of $\phi$ and a copy of the identity. Therefore
$$ 4 = d(\phi)^2 = d(\phi)\cdot d(\bar{\phi}) = d(\bar{\phi} \circ \phi) \geq d(id \oplus \phi \oplus \phi) = 1 + 2d(\phi) = 5 $$
which is a contradiction. Note that if you drop the assumption that $\phi$ is irreducible, there should be examples: Suppose $M$ carries an involution $\alpha \colon M \to M$, i.e. an action of $\mathbb{Z} / 2\mathbb{Z}$. Consider $\phi = id \oplus \alpha$ with the definition of the sum similar to the one in your question. Then
$$ [\phi \circ \phi] = [id \oplus \alpha] \circ [id \oplus \alpha] = [id \oplus \alpha \oplus \alpha \oplus \alpha^2] = [id \oplus \alpha \oplus id \oplus \alpha] = [\phi \oplus \phi] $$
where the brackets mean unitary equivalence classes of endomorphisms.
