Let $F$ be a finite extension of $\mathbf{Q}_p$ with integers $\mathscr{O}$, let $\mathbb{G}$ be a connected reductive group over $F$ and let $G=\mathbb{G}(F)$ be its $F$-points. Let $X(G)=\operatorname{Hom}(G,\operatorname{GL}(1))$ denote the group of algebraic characters of $G$ and let $G^0$ denote the intersection of $\chi^{-1}(\mathscr{O}^\times)$ as $\chi$ runs over $X(G)$. For example, if $G=\operatorname{GL}(n,F)$ then $G^0$ is the elements of $G$ with determinant equal to a unit. Informally, $G^0$ is generated by the derived subgroup of $G$ and the maximal compact subgroup of the centre of $G$.

Let $\pi$ be an irreducible supercuspidal representation of $G$ (over the complexes, as usual) and consider the restriction of $\pi$ to $G^0$. If $Z$ is the centre of $G$ then $G^0Z$ is a normal subgroup of finite index in $G$, and $G/G^0Z$ is abelian. Hence the restriction of $\pi$ to $G^0$ can be written as a finite direct sum of irreduibles $\pi_i$.

My question is: is there an example known where the isomorphism class of some $\pi_i$ occurs with multiplicity greater than one in $\pi$? That is—is any supercuspidal $\pi$ always "multiplicity-free" as a representation of $G^0$?

Why do I want to know this?

Those who know about Bernstein components will know the motivation behind this question.
When analysing the cuspidal Bernstein component of the category of smooth representations of $G$ associated to $\pi$, life can be very easy (Schur orthogonality relations etc) *if* all matrix coefficients of $\pi$ are actually compactly-supported. But a general connected reductive group may have a non-trivial centre, whose $F$-points are typically non-compact, which means compactly-supported matrix coefficients are very rare. The point about $G^0$ is that it has compact centre so these problems go away, but the representation theory of $G^0$ is very close to that of $G$. However, when moving from the study of the representation theory of $G^0$ to $G$ one has to induce back up; a typical ring that shows up in this procedure is $R:=\operatorname{End}_G(\operatorname{Ind}_{G^0}^G(\pi_i))$ for a $\pi_i$ as in the question. If $\pi_i$ shows up (up to isomorphism) with multiplicity greater than one then this ring will be non-commutative (in fact this is an iff). The Bernstein component corresponding to $\pi$ is isomorphic to the category of right $R$-modules, and if $R$ is commutative then it seems to me that it's always isomorphic to the algebraic functions on a product of $\operatorname{GL}(1,\mathbf{C})$'s so it's a really "easy" ring, but the literature seems to leave open the possibility that non-commutative $R$s can occur.

What do I know?

If $G/G^0$ is isomorphic to the integers then multiplicities greater than one cannot occur. In particular if $G=GL_n$ then multiplicities greater than one cannot occur. If $G$ is a product of $GL_n$'s then the same arguments apply and multiplicities greater than one cannot occur. If $\pi$ admits a Whittaker model then $\pi$ is multiplicity-free as a representation of $G_{der}$ and hence as a representation of $G^0$, and so again multiplicites cannot occur. I learnt this from Remark 1.6.1.3 of "The Bernstein decomposition and the Bernstein centre" by Alan Roche. Beyond this I know nothing.