The answer to the question is no, $T$ is not co-Hopfian, i.e., it does contain proper subgroups isomorphic to itself. Nicolás Matte Bon explained this to me over email (he doesn't use Mathoverflow, but someone showed him this question).

Matte Bon's strategy is to look at $T$ acting on the Cantor set $C=\{0,1\}^{\mathbb{N}}$ and define a new (faithful) action of $T$ on $C$ that yields an (injective) endomorphism $T\to T$ (and then argue that it is not surjective). For non-surjectivity, the idea is that under the new action there is a proper non-empty invariant open subset (so the action is not minimal), unlike for the usual action of $T$. Constructing such an action is a bit too complicated to explain here, but it uses ideas from Section 11 of Matte Bon's paper [1]. What I can do here is explain one concrete example of a proper endomorphism $T\to T$, which I sorted out after understanding Matte Bon's general construction.

To describe the endomorphism $T\to T$, I will use the strand diagram model for elements of $T$ (and $F$ and $V$), see, e.g., Definition 2.7 of Belk-Matucci [2]. For each split or merge vertex (Definition 2.1(2)), draw a small neighborhood around the vertex, not meeting any other vertices. For a split this neighborhood has one incoming strand and two outgoing strands, and for a merge this neighborhood has two incoming strands and one outgoing strand. Now to define $T\to T$ we replace the picture inside each such neighborhood with a more complicated picture. For a split, replace the strand going from the split vertex to the right exit with the picture in Figure 2 of Belk-Matucci (the usual "$x_0$" generator). For a merge, do this same thing but flipped upside-down. (Don't change any cyclic permutations.) This defines a well defined injective endomorphism $T\to T$ (if we only allow cyclic permutations that is; if we allow all permutations then it's $V\to V$, and if we allow no permutations then it's $F\to F$). To see it's non-surjective, one can use the strand diagram method of analyzing dynamics (see, e.g., Figure 19 of Belk-Matucci) to check that if $f$ is in the image of this endomorphism and $c\in C$ starts with $11$ then $f(c)$ has a "$11$" somewhere in it. (Since $T$ certainly contains elements violating this rule, this gives non-surjectivity.)

[1] Nicolás Matte Bon, *Rigidity properties of full groups of pseudogroups over the Cantor set*. arXiv link

[2] James Belk, Francesco Matucci, *Conjugacy and dynamics in Thompson's groups.* Geometriae Dedicata 169.1 (2014) 239-261. arXiv link

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