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. The automorphism group $k$-scheme of $(X,\mathcal{L})$, $\text{Aut}_k(X,\mathcal{L})$, is an affine group $k$-scheme. This is discussed many places, e.g., in my article with de Jong on discriminant avoidance.
Let $T$ be a (geometrically) reductive group $k$-scheme. For every homomorphism of group $k$-schemes, $$\rho:T\to \text{Aut}_k(X,\mathcal{L}),$$ define the semistable locus in the usual way, $$X^{\text{ss}}_\rho:=\cup\{ D_{\mathcal{L}^{\otimes n}}(s) | n\in \mathbb{Z}_{>0}, \ s \in H^0(X,\mathcal{L}^{\otimes n})^{\rho(T)}\}.$$ I hope that this rather pedantic formulation makes clear that the semistable locus depends only on the image of $\rho$. Moreover, the set is also invariant under replacing $\rho$ by its composition with the standard homomorphism of affine group $k$-schemes, $$\gamma_n:\text{Aut}_k(X,\mathcal{L})\to \text{Aut}_k(X,\mathcal{L}^{\otimes n}),$$ for each positive integer $n$.
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 $T$.