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. 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 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. 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. 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. 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 affine $k$-morphisms from $X^{\text{ss}}_\rho$ to $k$-schemes, there exists an initial such $k$-morphism, $$q:X^{\text{ss}}_\rho \to Y.$$ 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 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.
Corollary The GIT quotient depends only on the open subscheme $X^{\text{ss}}_\rho$ and the closed subscheme $R_\rho$ of $X^{\text{ss}}_\rho\times_{\text{Spec}\ k}X^{\text{ss}}_\rho.$
There is an induced action on $X^{\text{ss}}_\rho$ by the subgroup scheme $\rho(T)\subset \text{Aut}_k(X,\mathcal{L})$. The main theorem of Geometric Invariant Theorem asserts the existence of a geometric quotient of this induced action, $$q:X^{\text{ss}}_\rho \to Y.$$ This quotient morphism is affine, it is universally submersive, it is a uniform categorical quotient, etc. To make this very intrinsic, the quotient $q$ is the same as the quotient of $X^{\text{ss}}_\rho$ by the relation $R_q$ that is the closed subscheme of the self-product, $$R_q=X^{\text{ss}}_\rho\times_{q,Y,q} X^{\text{ss}}_\rho \subset X^{\text{ss}}_\rho\times_{\text{Spec}\ k} X^{\text{ss}}_\rho.$$ The geometric quotient is the uniform categorical quotient of the subscheme $X^{\text{ss}}_\rho$ by the relation that is the closed subscheme $R_q\subset X^{\text{ss}}_\rho\times_{\text{Spec}\ k}X^{\text{ss}}_\rho$.
Once we know the open subscheme $X^{\text{ss}}_\rho$ and the closed subscheme $R_q$, we are free to "forget" about the group actions -- the quotient is the uniform categorical quotient of $X^{\text{ss}}_\rho$ by $R_q$. As above: 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 $\gamma_n$. In this pedantically precise sense, the GIT quotient does not depend on "how we name" our reductive group $T$. (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.)
Here is my interpretation of the question by the OP.
Question. 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 with respect to the ample invertible sheaf $\omega_{X_{r,n}}^\vee$, for the associated geometric quotient $Y_{r,n}=X_{r,n}^{\text{ss}}/\rho(T)$, is there a $k$-isomorphism of $Y_{r,n}$ with $Y_{n-r,n}$?
Notice that I did not record the maximal torus in the notation for $Y_{r,n}$. For any other maximal torus $T',$ there exists an element $g$ such that $T'$ equals $gTg^{-1}$. Then the $k$-isomorphism, $$r_g:X_{r,n} \to X_{r,n}, \ \ x \mapsto x\cdot g^{-1},$$ intertwines $T$ and $T'$. Thus, $r_g$ restricts to an isomorphism, $$X^{\text{ss}}_{r,n,T} \to X^{\text{ss}}_{r,n,T'}.$$ Moreover, conjugation by $g$ intertwines the the $T$-action and $T'$-action with respect to $r_g$, i.e., $r_g$ is $T$-equivariant if we define the $T$-action on $X^{\text{ss}}_{r,n,T'}$ to act through conjugation by $g$. Stated differently, the induced map, $$(r_g,r_g):X^{\text{ss}}_{r,n,T}\times_{\text{Spec}\ k} X^{\text{ss}}_{r,n,T} \to X^{\text{ss}}_{r,n,T'}\times_{\text{Spec}\ k} X^{\text{ss}}_{r,n,T'},$$ maps the relation $R_T$ to the relation $R_{T'}$. Thus, there is an induced isomorphism of the geometric quotients. It is in this sense that the geometric quotient of $\text{Grass}(r,n)$ is independent of the choice of maximal torus; the quotient by $T$ and the quotient by $T'$ are isomorphic as $k$-schemes.
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 $\phi$, $$\phi( gHg^{-1}) = \phi(g)\phi(H)\phi(g)^{-1}.$$ Thus, for the conjugation action of a maximal torus $T$ on $P_r$, for the image maximal torus $T'=\phi(T)$, the induced $k$-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 $R_T$ to the closed subscheme $R_{T'}$.
There are many ways to be confused at this moment, but I hope the last two sentences above are unambiguous. Since the respective geometric quotients are uniform categorical quotients of $(P_r)^{\text{ss}}_T$, resp. of $(P_{n-r})^{\text{ss}}_{T'}$, by the relation that is the closed subscheme $R_T$ of the self-product, resp. of the closed subscheme $R_{T'}$ of the self-product, and since $\phi_r$ induces an isomorphism between these, by the universal property of a uniform categorical quotient, there is an induced isomorphism of the uniform categorical quotients.
I want to emphasize this once more: I am not claiming that $\phi_r$ is equivariant for the standard action of $\textbf{SL}_n$ on $P_r$ and the standard action of $\textbf{SL}_n$ on $P_{n-r}$. It is equivariant if we intertwine the actions. Since $\phi$ maps the maximal torus $T$ to a maximal torus $T'$, also $\phi$ induces a $k$-isomorphism $P_r//T \to P_{n-r}//T'$ in the precise version of the previous paragraph. Finally, as explained earlier, for $g\in \textbf{SL}_n$ that conjugates $T$ and $T'$ (and this is very non-unique), conjugation by $g$ induces an isomorphism between $P_{n-r}//T'$ and $P_{n-r}//T$. Thus, there is a $k$-isomorphism of $P_r//T$ and $P_{n-r}//T$, even though there is no group $k$-scheme automorphism $\phi_r$ of $\textbf{SL}_n$ that restricts to the identity map from $T$ to $T$.