I'm rushed, so this might not be right, but it feels right. Consider reduced operads in a cartesian monoidal category, so that $C(0) = \ast$. A monoid $M$ in our category gives a reduced operad $R(M)$ with $j$th object $M^j$. The structure map $$M^k\times M^{j_1}\times\cdots M^{j_k}\to M^{j_1+\cdots + j_k}$$
has coordinates given by multiplying the $h$th coordinate of $M^k$ with the coordinates of $M^{j_h}$. This is a right adjoint to the functor $L$ that sends a reduced operad $C$ to $C(1)$ with the evident product and unit. The isomorphism of hom sets uses the degeneracy operators that are there because we are working with reduced operads. Obviously $LR= id$. Very often $RL$ is an inclusion. That happens for little whatnots. Very rarely, only in uninteresting cases, is $id\to RL$ an isomorphism. That doesn't happen for little whatnots.

I'm rushed, so this might not be right, but it feels right. Consider reduced operads in a cartesian monoidal category, so that $C(0) = \ast$. A monoid $M$ in our category gives a reduced operad $R(M)$ with $j$th object $M^j$. The structure map $$M^k\times M^{j_1}\times\cdots M^{j_k}\to M^{j_1+\cdots + j_k}$$
has coordinates given by multiplying the $h$th coordinate of $M^k$ with the coordinates of $M^{j_h}$. This is a right adjoint to the functor $L$ that sends a reduced operad $C$ to $C(1)$ with the evident product and unit. The isomorphism of hom sets uses the degeneracy operators that are there because we are working with reduced operads. Obviously $LR= id$. Very often $id\to RL$ is an inclusion. That happens for little whatnots. Very rarely, only in uninteresting cases, is $id\to RL$ an isomorphism. That doesn't happen for little whatnots.

I'm rushed, so this might not be right, but it feels right. Consider reduced operads in a cartesian monoidal category, so that $C(0) = \ast$. A monoid $M$ in our category gives a reduced operad $R(M)$ with $j$th object $M^j$. The structure map $$M^k\times M^{j_1}\times\cdots M^{j_k}\to M^{j_1+\cdots + j_k}$$
has coordinates given by multiplying the $h$th coordinate of $M^k$ with the coordinates of $M^{j_h}$. This is a right adjoint to the functor $L$ that sends a reduced operad $C$ to $C(1)$ with the evident product and unit. The isomorphism of hom sets uses the degeneracy operators that are there because we are working with reduced operads. Obviously $LR= id$. Very often $RL$ is an inclusion. That happens for little whatnots. Very rarely, only in uninteresting cases, is $RL$ an isomorphism. That doesn't happen for little whatnots.

I'm rushed, so this might not be right, but it feels right. Consider reduced operads in a cartesian monoidal category, so that $C(0) = \ast$. A monoid $M$ in our category gives a reduced operad $R(M)$ with $j$th object $M^j$. The structure map $$M^k\times M^{j_1}\times\cdots M^{j_k}\to M^{j_1+\cdots + j_k}$$
has coordinates given by multiplying the $h$th coordinate of $M^k$ with the coordinates of $M^{j_h}$. This is a right adjoint to the functor $L$ that sends a reduced operad $C$ to $C(1)$ with the evident product and unit. The isomorphism of hom sets uses the degeneracy operators that are there because we are working with reduced operads. Obviously $LR= id$. Very often $RL$ is an inclusion. That happens for little whatnots. Very rarely, only in uninteresting cases, is $id\to RL$ an isomorphism. That doesn't happen for little whatnots.