Let $\mathbf{Poly}$ denote the category of polynomial functors on $\mathbf{Set}$, and let $\mathfrak{m}\colon\mathbf{Poly}\to\mathbf{Poly}$ be the free monad monad, i.e. the functor that sends every polynomial $p$ to the free monad $(\mathfrak{m}_p,\eta,\mu)$,
$$
\mathfrak{m}_p\colon\mathbf{Set}\to\mathbf{Set}
\qquad
\eta_p\colon\text{id}_\mathbf{Set}\to \mathfrak{m}_p
\qquad
\mu_p\colon \mathfrak{m}_p\circ \mathfrak{m}_p\to \mathfrak{m}_p
$$
on $p$. One can think of the underlying functor $\mathfrak{m}_p$ as the coproduct over all $p$-syntax trees of the functor represented by the leaves. 

Let $(t,\eta,\mu)$ denote an arbitrary polynomial monad on $\mathbf{Set}$. This means $t$ comes equipped with an algebra structure $\alpha\colon\mathfrak{m}_t\to t$ satisfying a unit law $\alpha\circ\eta_t=\text{id}_t$ and a similar composition law. We define a *self-descriptor of $t$* to be a section
$$
\beta\colon t\to\mathfrak{m}_t$$
of the algebra map, $\alpha\circ\beta=\mathrm{id}_t$. We say that the self-descriptor $\beta$ is *trivial* if $\beta=\eta_t$, and we say it is *nontrivial* if $\beta\neq\eta_t$.

Note that a free monad $\mathfrak{m}_p$ always has a nontrivial self-descriptor, namely
$$
\mathfrak{m}_p\xrightarrow{\mathfrak{m}_{\eta_{\,p}}}\mathfrak{m}_{\mathfrak{m}_p}
$$
which is indeed nontrivial because $\mathfrak{m}_{\eta_p}\neq\eta_{\mathfrak{m}_p}$.

**Question:** 
If a polynomial monad $t$ on $\mathbf{Set}$ has a non-trivial self-descriptor, does that imply that $t$ is free, $t=^?\mathfrak{m}_p$ on some polynomial $p$? Or, instead, does there exist a non-free monad that has a non-trivial self-descriptor?