I am writing this as an answer because the comments are already too long. In the following I am incredibly pedantic, because there seems to be endless possibility for confusion with the several simultaneous group schemes, group scheme elements, and group scheme automorphisms that are involved.
Let $k$ be a field; later I will assume that $k$ is algebraically closed. Let $X$ be a projective $k$-scheme. Let $\mathcal{L}$ be an ample invertible sheaf on $X$. For instance, if $X$ is a Fano manifold (such as any projective homogeneous space), we might choose $\mathcal{L}$ to be the determinant of the tangent bundle.
Lemma 1. The automorphism group $k$-scheme of $(X,\mathcal{L})$, $\text{Aut}_k(X,\mathcal{L})$, is an affine group $k$-scheme.
Proof. This is discussed many places, such as Subsection 2.1 of the following article.
A. J. de Jong and J. Starr
Almost proper GIT-stacks and discriminant avoidance
Doc. Math. 15 (2010), pp.957-972.
I am pretty sure that this is also in Grothendieck's Bourbaki seminar notes, "Fondements de la Géométrie Algébrique." QED
Let $T$ be a (geometrically) reductive group $k$-scheme. Let $$\rho:T\to \text{Aut}_k(X,\mathcal{L})$$ be a morphism of group $k$-schemes.
Definition 2 The semistable locus of $\rho$ is the open subscheme of $X,$ $$X^{\text{ss}}_\rho:=\cup\{ D_{\mathcal{L}^{\otimes n}}(s) | n\in \mathbb{Z}_{>0}, \ s \in H^0(X,\mathcal{L}^{\otimes n})^{\rho(T)}\}.$$
For every integer $n$, denote by $\gamma_n$ the natural morphism of group $k$-schemes, $$\gamma_n:\text{Aut}_k(X,\mathcal{L})\to \text{Aut}_k(X,\mathcal{L}^{\otimes n}).$$
Lemma 3. The open subscheme $X^{\text{ss}}_\rho$ of $X$ depends only on the image of $\rho$. Also, for every integer $n$, $X^{\text{ss}}_{\gamma_n\circ \rho}$ equals $X^{\text{ss}}_{\rho}$ as open subschemes of $X$.
Proof. The first statement is straightforward from the definition of the semistable locus, which only depends on the $\rho(T)$. For the second statement, for every $\rho(T)$-invariant section $s$, also $s^n$ is invariant under $\gamma_n\circ\rho(T)$. Since $D(s)$ equals $D(s^n)$, it follows that the open subschemes are equal. QED
Definition 4. The associated relation $R_\rho$ is the closed image subscheme (i.e., minimal closed subscheme through which the morphism factors) of the morphism $$\Psi_\rho:T\times_{\text{Spec}\ k} X^{\text{ss}}_{\rho} \to X^{\text{ss}}_\rho \times_{\text{Spec}\ k} X^{\text{ss}}_\rho, \ \ (t,x)\mapsto (\rho(t,x),x).$$
Lemma 5. A separated $k$-morphism $q:X^{\text{ss}}\to Y$ is invariant for the action $\rho$ if and only if the fiber product $X^{\text{ss}}\rho \times_{q,Y,q}X^{\text{ss}}\rho$ contains $R_\rho$ as closed subschemes of $X^{\text{ss}}_\rho\times_{\text{Spec}\ k}X^{\text{ss}}_{\rho}.$
Proof. By the definition of $R_\rho$, the fiber product contains $R_\rho$ if and only if it contains the image of $\psi_\rho$. QED
Fundamental Theorem of Geometric Invariant Theory (Projective Case). [Theorem 1.10, p. 38, Geometric Invariant Theory, 3rd ed., Mumford, Fogarty, Kirwan, Ergebnis. Math. 34, Spring-Verlag, Berlin, 1994]
Among all $G$-invariant $k$-morphisms from $X^{\text{ss}}_\rho$ to $k$-schemes, there exists an initial such $k$-morphism, $$q:X^{\text{ss}}_\rho \to Y,$$ i.e., $q$ is a categorical quotient. This morphism is a uniform categorical quotient, i.e., it is a categorical quotient after finitely presented, flat base change of $Y$. This morphism is affine. This morphism is universally submersive, i.e., for every finitely presented morphism $Y'\to Y$ and for every subset $U\subset Y$, $U$ is open if and only if the inverse image of $U$ in $Y'\times_{Y,q} X^{\text{ss}}_\rho$ is open. The scheme $Y$ is projective, and there exists an ample invertible sheaf whose pullback under $q$ with its natural linearization is isomorphic to a positive tensor power of $\mathcal{L}$ with its induced linearization. Finally, there exists an open subscheme $Y_0$ of $Y$ whose inverse image equals the (properly) stable locus $X^{\text{s}}_{\rho,0}$, and the restriction $q_0:X^{\text{s}}_{\rho,0}\to Y_0$ is a uniform geometric quotient.
Let $(X,\mathcal{L})$ and $(X',\mathcal{L}')$ be polarized projective $k$-schemes. Let $T$ and $T'$ be reductive group $k$-schemes. Let $$\rho:T\to \text{Aut}_k(X,\mathcal{L}), \ \ \rho:T'\to \text{Aut}_k(X',\mathcal{L}'),$$ be morphisms of group $k$-schemes. Denote the GIT quotients above by $$q:X^{\text{ss}}_\rho\to Y, \ \ q':(X')^{\text{ss}}_{\rho'}\to Y'.$$
Corollary 6. For every $k$-morphism, $\phi:X\to X',$ if the open subscheme $\phi^{-1}(X')^{\text{ss}}_{\rho'}$ of $X$ contains $X^{\text{ss}}_\rho$, and if the closed subscheme $\phi^{-1}(R_{\rho'})\cap (X^{\text{ss}}_\rho\times_{\text{Spec}\ k}X^{\text{ss}}_\rho)$ contains $R_\rho$, then there exists a unique $k$-morphism $\phi_Y:Y\to Y'$ such that $q'\circ \phi$ equals $\phi_Y\circ q$.
Proof. By the first hypothesis, $q'\circ\phi$ is defined as a $k$-morphism $X^{\text{ss}}_\rho\to Y'$. By the second hypothesis, $q'\circ \phi$ is invariant for the action $\rho$. Thus, since $q$ is a categorical quotient, there exists a unique $k$-morphism $\phi_Y$ such that $\phi_Y\circ q$ equals $q'\circ\phi$. QED
As a special case, note that the open subscheme $X^{\text{ss}}_\rho\subset X$ and the closed subscheme $R_q\subset X^{\text{ss}}_\rho\times_{\text{Spec}\ k}X^{\text{ss}}_\rho$ depend only on the image of $\rho$, and they are also invariant under composing $\rho$ with $\gamma_n$. Thus, the open subscheme $X^{\text{ss}}_\rho$ and the $k$-morphism $q:X^{\text{ss}}_\rho\to Y$ depend only on the image of $\rho$, and they are invariant under composing $\rho$ with $\gamma_n$. (I am not saying that there are no other interesting structures coming from the group action. I am only saying that the quotient morphism depends only on the semistable locus and the relation.)
A more substantial example is as follows. Let $G$ be a reductive group $k$-scheme, and let $\rho$ be a morphism of group $k$-schemes, $$\rho:G\to \text{Aut}_k(X,\mathcal{L}).$$ For each closed subgroup $k$-scheme $T$ of $G$ that is reductive, denote by $\rho_T$ the restriction morphism of group $k$-schemes, $$\rho_T:T\to \text{Aut}_k(X,\mathcal{L}).$$ Let $T$ and $T'$ be closed subgroup $k$-schemes of $G$ that are both reductive. Denote the uniform categorical quotient $k$-morphisms by, $$q:X^{\text{ss}}_{\rho_T} \to Y_T, \ \ q':X^{\text{ss}}_{\rho_{T'}} \to Y_{T'}.$$ Let $g$ be a $k$-point of $G$ such that the conjugation, $$c_g:G\to G, \ \ h\mapsto ghg^{-1},$$ restricts to an isomorphism from $T$ to $T'$. Denote by $r_g$ the associated right translation, $$r_g:X \to X, \ \ x \mapsto x\cdot \rho(g),$$ with its associated isomorphism $r_g^*\mathcal{L}\cong \mathcal{L}$.
Corollary 7. The $k$-isomorphism $r_g$ mapso $X^{\text{ss}}_{\rho_T}$ isomorphically to $X^{\text{ss}}_{\rho_{T'}}$. The inverse image of the relation $R_{T'}$ equals the relation $R_T$. There is a unique $k$-morphism $r_{g,Y}:Y_T\to Y_{T'}$ such that $r_{g,Y}\circ q$ equals $q'\circ r_g$.
Proof. This is a special case of the previous corollary. QED
Question 8. Let $k$ be an algebraically closed field. Let $T$ be a maximal torus in $\textbf{SL}_n$. For the natural right action of $\textbf{SL}_n$ on the Grassmannian $X_{r,n}=\textbf{Grass}(r,n)$, with its unique linearization of the ample invertible sheaf $\omega_{X_{r,n}}^\vee$, for the associated geometric quotient $Y_{r,n,T}=X_{r,n}^{\text{ss}}/\rho(T)$, is there a $k$-isomorphism of $Y_{r,n,T}$ with $Y_{n-r,n,T}$?
For each $k$-point of $\text{Grass}(r,n)$, there is an associated $\textbf{SL}_n$-equivariant isomorphism of $\text{Grass}(r,n)$ with the $k$-scheme $P_r$ parameterizing parabolic subgroup schemes $H$ of $\textbf{SL}_n$ in the same conjugacy class as the stabilizer of the $k$-point. The induced action of $\textbf{SL}_n$ on $P_r$ is by conjugation.
There exists an outer automorphism (very much not unique), $$\phi:\textbf{SL}_n \to \textbf{SL}_n,$$ that sends every parabolic parameterized by $P_r$ to a parabolic parameterized by $P_{n-r}$. Thus, there is an induced $k$-isomorphism, $$\phi_r: P_r \to P_{n-r}, \ \ H \mapsto \phi(H).$$ This $k$-isomorphism is certainly not $\textbf{SL}_n$-equivariant, rather the two actions are intertwined by $\phi$, $$\phi( gHg^{-1}) = \phi(g)\phi(H)\phi(g)^{-1}.$$ For each maximal torus $T$ in $\textbf{SL}_n$, denote by $T'$ the image maximal torus $\phi(T').$ Denote the uniform categorical quotients as follows, $$q:(P_r)^{\text{ss}}_T \to Y_T, \ \ q':(P_{n-r})^{\text{ss}}_{T'} \to Y'_{T'}.$$
Corollary 9. The $k$-isomorphism $\phi_r$ restricts to a $k$-isomorphism, $$(P_r)^{\text{ss}}_T \to (P_{n-r})^{\text{ss}}_{T'}.$$ Moreover, the product $k$-isomorphism, $$(\phi_r,\phi_r):(P_r)^{\text{ss}}_T\times_{\text{Spec}\ k} (P_r)^{\text{ss}}_T \to (P_{n-r})^{\text{ss}}_{T'}\times_{\text{Spec}\ k} (P_{n-r})^{\text{ss}}_{T'},$$ maps the closed subscheme $R_T$ isomorphically to the closed subscheme $R_{T'}$. There is a unique $k$-morphism $\phi_{r,Y}:Y_T\to Y'_{T'}$ such that $\phi_{r,Y}\circ q$ equals $q'\circ \phi_r.$
Proof. The follows in the same way as the previous two corollaries. QED
Proposition 10. There exists a $k$-isomorphism of $Y_{r,n,T}$ with $Y_{n-r,n,T}.$
Proof. By the first of the two corollaries above, there exists a $k$-isomorphism of $Y_{n-r,n,T}$ with $Y_{n-r,n,T'}.$ By the second corollary, there exists a $k$-isomorphism of $Y_{r,n,T}$ with $Y_{n-r,n,T'}.$ QED