3 Corrected proof and added more detail.

Proof. Since $\pi$ preserves (ii) we have that $\pi(1_S 1_T) = \pi(1_S) \wedge \pi(1_T)$, for all $S, T \subset X$. Since $\pi(1_X) = 1$ and since $\pi$ preserves (iv) we have that $\pi( 1_X - 1_S ) = 1 - \pi(1_S)$ for all $S \subset X$. Hence by again using the fact that $\pi$ preserves (iv) we have that $\pi( 1_{S \cup T}) = \pi( 1_S ) \vee \pi( 1_T)$ andA standard fact from operator algebras then shows that $\pi(1_T)$ and $\pi(1_S)$ commute. Indeed, if we consider the polar decomposition left hand side $(1 \pi(1_S) - \pi(1_S) \wedge \pi(1_T)$ is the projection onto the closure of the range of $\pi(1_S)(1 - \pi(1_T))$ and satisfies \pi(1_S) \wedge (1 - \pi(1_T)) \leq \pi(1_S) = VA$then$VV^* = - \pi(1_S) \wedge \pi(1_T) \leq \pi(1_S),$$while the right hand side \pi(1_S) \vee \pi(1_T) - \pi(1_T) = pi(1_T) is the projection onto the closure of the range of (1 - \pi(1_T))\pi(1_S) and satisfies \pi(1_S) \wedge (1 - \pi(1_T)) \leq \pi(1_S) \wedge vee \pi(1_T) = V^*V, hence - \pi(1_T) \leq 1 - \pi(1_T).V is a projection,  Hence both sides equal P = \pi(1_S) \wedge (1 - \pi(1_T)) and so we have(1 \pi(1_S)(1 - \pi(1_T)) = P \pi(1_S) is self-adjoint pi(1_S)(1 - \pi(1_T))= P = P (1 - \pi(1_T))\pi(1_S) = (1 - \pi(1_T))\pi(1_S), 2 Fixed typo. = \pi(1_{S \cup T} - 1_{S} 1_T) = \pi(1_S) \vee \pi(1_T) - \pi(1_S).A standard fact from operator algebras then shows that \pi(1_T) and \pi(1_S) commute. Indeed, if we consider the polar decomposition (1 - \pi(1_T)) \pi(1_S) = VA then VV^* = \pi(1_S) \vee \pi(1_T) - \pi(1_S) pi(1_T) = \pi(1_S) - \pi(1_S) \wedge \pi(1_T) = V^*V, hence V is a projection, and so (1 - \pi(1_T)) \pi(1_S) is self-adjoint which shows that \pi(1_T) and \pi(1_S) commute. Since S and T were arbitrary this gives the result. \square 1 After looking at the paper of Ceccherini-Silberstein, Grigorchuk, and de la Harpe linked to in the question I think I have a better idea of what is being asked, so let me record some further thoughts here in a separate answer. Given a set X and a bijection \gamma : S \to T between subsets of X let me denote by s(\gamma) = S (the support) and r(\gamma) = T (the range). As in the above paper let me define a (standard) pseudogroup \mathcal G of transformations on X to be a set of bijections \gamma : S \to T between subsets S, T \subset X such that: (i) {\rm id}_{|S} \in \mathcal G, for any subset S \subset X. (ii) If \gamma \in \mathcal G then \gamma^{-1} \in \mathcal G. (iii) If \gamma, \delta \in \mathcal G then \delta \circ \gamma \in \mathcal G, where \delta \circ \gamma is understood to have domain \gamma^{-1}(r(\gamma) \cap s(\delta)). (iv) If \gamma: S \to T is a bijection of subsets S, T \subset X and S = \sqcup_{1 \leq j \leq n} S_j is a finite partition with \gamma_{|S_j} \in \mathcal G for all 1 \leq j \leq n, then \gamma \in \mathcal G. I will only consider the case when X a discrete space with counting measure, although much of this should work in the setting where X is is a standard Borel space with a (possibly infinite) measure \lambda on X such that \lambda(s(\gamma)) = \lambda(r(\gamma)) for all \gamma \in \mathcal G. Let me make now a "quantum" definition. A pseudogroup \mathcal H of quantum transformations on a Hilbert space K is a set of partial isometries on K such that: (i) 1 = {\rm id}_{K} \in \mathcal H, and the von Neumann algebra generated by the set of projections in \mathcal H contains a maximal abelian self-adjoint subalgebra (MASA) of \mathcal B(K). (ii) If v \in \mathcal H then v^* \in \mathcal H. (iii) If v, w \in \mathcal H then w \cdot v \in \mathcal H where w \cdot v is the partial isometry w( w^*w \wedge vv^* ) v, e.g., if v is a partial isometry from K_1 to K_2 and w is a partial isometry from K_3 to K_4 then w \cdot v is a partial isometry from v^*( K_2 \cap K_3 ) to w( K_2 \cap K_3). (iv) If v \in \mathcal B(K) is a partial isometry and v^*v = \Sigma_{1 \leq j \leq n} p_j is a finite partition of projections with v p_j \in \mathcal H for all 1 \leq j \leq n, then v \in \mathcal H. A nice example is the set of all partial isometries S(K) described by Mark above. A pseudogroup of transformations on a set X has a natural representation (which I will call the regular representation) as a pseudogroup of quantum transformations on \ell^2X, by viewing a bijection of subsets \gamma: S \to T as the partial isometry v_\gamma : \ell^2S \to \ell^2T given by v_\gamma(\xi) = \xi \circ \gamma^{-1}. Conversely, if \mathcal H is a pseudogroup of quantum transformations on K such that the set of projections in \mathcal H generates a purely atomic MASA then by letting X be the set of rank one projections in \mathcal H, we have a natural identification K = \ell^2X and we have that each v \in \mathcal H induces a bijection \gamma : S \to T of subsets of X such that v = v_\gamma. A homomorphism between (quantum) pseudogroups \mathcal G and \mathcal H is a map \pi: \mathcal G \to \mathcal H which preserves the structures (i)-(iv). If both \mathcal G and \mathcal H are pseudogroups of transformations on X and Y then homomorphisms are somewhat rigid, if we restrict to characteristic functions on points we obtain a map f: X \to Y and we then can check that \pi(\gamma) \circ f = f \circ \gamma, for all \gamma \in \mathcal G. Given a homomorphism between standard pseudogroups \mathcal G and \mathcal H we obtain a representation of \mathcal G by composition with the regular representation for \mathcal H. One could hope that by viewing a pseudogroup of transformatins \mathcal G as a pseudogroup of quantum transformations then maybe the homomorphism (and hence also the representation) structure would be richer. This however does not give a larger class. Proposition. Let \mathcal G be a pseudogroup of transformations on a set X. If \pi: \mathcal G \to S(K) is a representation then \pi(2^X) generates an abelian von neumann subalgebra of \mathcal B(K), hence every representation of \mathcal G is given by composing a homomorphism into a standard pseudogroup with the regular representation. Proof. Since \pi preserves (ii) we have that \pi(1_S 1_T) = \pi(1_S) \wedge \pi(1_T), for all S, T \subset X. Since \pi(1_X) = 1 and since \pi preserves (iv) we have that \pi( 1_X - 1_S ) = 1 - \pi(1_S) for all S \subset X. Hence by again using the fact that \pi preserves (iv) we have that$$ \pi(1_S) - \pi(1_S) \wedge \pi(1_T) = \pi(1_S - 1_{S \cap T})  = \pi(1_{S \cup T} - 1_{S} ) = \pi(1_S) \vee \pi(1_T) - \pi(1_S).  A standard fact from operator algebras then shows that $\pi(1_T)$ and $\pi(1_S)$ commute. Indeed, if we consider the polar decomposition $(1 - \pi(1_T)) \pi(1_S) = VA$ then $VV^* = \pi(1_S) \vee \pi(1_T) - \pi(1_S) = \pi(1_S) - \pi(1_S) \wedge \pi(1_T) = V^*V$, hence $V$ is a projection, and so $(1 - \pi(1_T)) \pi(1_S)$ is self-adjoint which shows that $\pi(1_T)$ and $\pi(1_S)$ commute. Since $S$ and $T$ were arbitrary this gives the result. $\square$

One could certainly define property (T) for pseudogroups of transformations in this setting by requiring that any representation which almost contains invariant vectors must contain a non-trivial invariant vectors. But given the above proposition it is equivalent to say that $\mathcal G$ has property (T) if and only if any quotient of $\mathcal G$ which is amenable must be finite. (Recall that a pseudogroup $\mathcal H$ of transformations on $X$ is amenable if there is a state $\varphi$ on $\ell^\infty X$, such that $\varphi(\gamma \gamma^{-1}) = \varphi(\gamma^{-1} \gamma)$, for all $\gamma \in \mathcal H$).

This definition seems like an interesting property, but it lacks a bit in functoriality. For instance, if $\Gamma$ is a countable group and we consider the pseudogroup $\mathcal G(\Gamma)$ of all maps on subsets of $\Gamma$ which arise from the action of $\Gamma$ on itself by left multiplication. Then I believe that the property that $\mathcal G(\Gamma)$ has no quotient onto an infinite amenable pseudogroup, can be restated in terms of group actions as: Every transitive action of $\Gamma$ on an infinite set $Y$ has no invariant mean.

If $\Gamma$ has property (T) then this condition is satisfied, but this condition will also be satisfied for groups which do not have property (T), although it seems to be non-trivial to show this. If $\Gamma$ is an infinite product of infinite simple property (T) groups then this should be an example. This condition is also satisfied for Tarski monsters (are there Tarski monsters which do not have property (T)?).

Coming to question 2, $I_n(\mathbb Z)$ in this setting is not a pseudogroup of transformations on a set, but rather a pseudogroup of quantum transformations in $M_n(\mathbb R) \subset M_n(\mathbb C)$. Finite dimensional algebras correspond to finite sets in the standard setting and so I think that $I_n(\mathbb Z)$ in this setting should be considered as "finite", and hence will have property (T) but for trivial reasons.

For instance, the definition of amenability for a pseudogroup $\mathcal H$ of quantum transformations on $H$ should be that there exists a state $\phi$ on $\mathcal B(K)$ such that $\phi(v^*v) = \phi(vv^*)$ for all $v \in \mathcal H$. This definition is consistent with the definition in the standard setting.

There may be another perspective in which $I_n(\mathbb Z)$ can be seen to be "rigid", but I'm not sure what it would be.

One could also fix a dimension $d$ and consider the action of $SL_n(\mathbb Z)$ on the Grassmannian $Gr(d, \mathbb R^n)$ and ask if the generated pseudogroup of transformations on $Gr(d, \mathbb R^n)$ does not have any non-trivial amenable quotients ($d \not= 0, n$). (This is no longer the discrete setting.) Using the property (T) of $SL_n(\mathbb Z)$ it should be enough to check the following property (perhaps this is well known or is discussed in the paper of Popa and Vaes that I referred to before, I am not sure): If $Y$ is a non-finite compact Hausdorff space which has a continous action of $SL_n(\mathbb Z)$ and such that there is a continuous $SL_n(\mathbb Z)$-invariant surjection $\pi: Gr(d, \mathbb R^n) \to Y$, can $Y$ have a $SL_n(\mathbb Z)$-invariant probability measure?