Recall that a monoidal monad on a monoidal category $(\mathcal{C}, \otimes, I)$ is a monad $(M, \eta, \mu)$ on $\mathcal{C}$ such that $M$ is also equipped with the structure of a lax monoidal functor in a way that makes $\eta$ and $\mu$ monoidal natural transformations.
The structure of a monoidal monad on a given monad is equivalent to the structure of a commutative monad: a pair of a left and right monad strengths $(\sigma, \tau)$ such that the two induced lax monoidal structures agree:
$$\mu \circ M\sigma \circ \tau = \mu \circ M\tau \circ \sigma : MA \otimes MB \to M (A \otimes B)$$
If furthermore $\mathcal{C}$ is a symmetric monoidal category with $\beta : A \otimes B \to B \otimes A$, then a monoidal monad is symmetric if the underlying monoidal functor is, and a strength is symmetric if $\tau = M\beta \circ \sigma \circ \beta$. Over the above equivalence, the property of being a symmetric monoidal monad is equivalent to the property of being a symmetric commutative monad (that is, a monad with a commutative symmetric strength). [1]
Question: is there a monoidal monad, on a symmetric monoidal category, that is not symmetric?
Equivalently: is there a commutative monad, on a symmetric monoidal category, whose left and right strengths are not related by the braiding?
On the Lambek Calculus with an Exchange Modality claims (lemma 9) to have a non-symmetric monoidal monad, but doesn't actually prove that it's non-symmetric — only that it need not be. The paper then concludes that the monad isn't commutative, which seems odd: being commutative is a property of a strength, not a monad, and since the monad is monoidal there is some commutative strength on it — just not the one you'd like. This all makes me wonder if I should trust the claim at all.
There is also a thread of people failing to answer this question at the nForum.
[1]: this is a decomposition of a well-known result by Kock; see 1Lab for an Agda formalisation of these claims.
Best Answer
[I crossposted this question on the category theory Zulip and got answers there, so I can now answer my own question; see there for more examples.]
An example of a non-symmetric commutative strong monad can be found at the very end of What makes a strong monad? (appendix A.2), which we can rephrase as a non-symmetric monoidal monad:
Consider a commutative monoid $(M, *)$, and let $\mathbf{Act} M$ be the category of sets equipped with a right $M$-action (that is, the functor category $[\mathbf{B}M, \mathbf{Set}]$), equipped with its cartesian monoidal structure inherited from $\mathbf{Set}$. The trivial action on $M$ (that is, the constant functor at $M$) is a monoid in $\mathbf{Act}M$, so we can talk about the writer monad $- \times M$.
This monad has a monoidal monad structure given by
\begin{align} \varphi_{A,B} &: (A \times M) \times (B \times M) \to (A \times B) \times M \\\\ &= ((a, m), (b, n)) \mapsto ((a, b \cdot m), m * n) \end{align}
where commutativity is used in checking compatibility with the monad's multiplication. This clearly isn't symmetric as long as $M$ is non-trivial, because
$$ ((a, b · m), m * n) \neq ((a · n, b), m * n) $$
in general.