This is too long to be a comment. But I felt it serves as a pointer.

Without loss of generality we assume $X_{1},X_{2},X_{3}$ share the same image domain $[0,1]$.

For a specific $\pi\in S_{3}$, the probability of $X_{\pi(1)}<X_{\pi(2)}<X_{\pi(3)}$ is $$\int_{0}^{1}dF_{X_{\pi(3)}}(\nu)\int_{0}^{\nu}dF_{X_{\pi(2)}}(\omega)\int_{0}^{\omega}dF_{X_{\pi(1)}}(\eta)=\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta$$where $F_{X_{i}}$ are probability measures on $[0,1]$; $f_{X_{i}}$ are probability densities w.r.t. Lebesgue measure on $[0,1]$, therefore it suffices to make the 3! integral equations have a consistent solution $f_{X_{1}},f_{X_{2}},f_{X_{3}}$. But since they are independent, $f_{X_{1}},f_{X_{2}},f_{X_{3}}$ are at the same time marginal probability measures, so the problem becomes finding a joint distribution $g$ of $(X_{1},X_{2},X_{3})$ subject to 3! conditions $$\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}f_{X_{\pi(3)}}(\nu)f_{X_{\pi(2)}}(\omega)f_{X_{\pi(3)}}(\eta)d\nu d\omega d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}g_{X_{\pi(3)},X_{\pi(2)},X_{\pi(1)}}(\nu,\omega,\eta)d\nu d\omega d\eta$$.

If you are willing to assume that $p_{\pi}\equiv\frac{1}{3!}$ then a special case that suffices to work is $X_{i}$ are exchangeable, which deserves an extended discussion like the one in Chap 5,7 in [Kallenberg].

[Kallenberg]Kallenberg, Olav. Probabilistic symmetries and invariance principles. Springer Science & Business Media, 2006.

This is too long to be a comment. But I felt it serves as a pointer.

Without loss of generality we assume $X_{1},X_{2},X_{3}$ share the same image domain $[0,1]$.

For a specific $\pi\in S_{3}$, the probability of $X_{\pi(1)}<X_{\pi(2)}<X_{\pi(3)}$ is $$\int_{0}^{1}dF_{X_{\pi(3)}}(\nu)\int_{0}^{\nu}dF_{X_{\pi(2)}}(\omega)\int_{0}^{\omega}dF_{X_{\pi(1)}}(\eta)=\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta$$where $F_{X_{i}}$ are probability measures on $[0,1]$; $f_{X_{i}}$ are probability densities w.r.t. Lebesgue measure on $[0,1]$, therefore it suffices to make the 3! integral equations have a consistent solution $f_{X_{1}},f_{X_{2}},f_{X_{3}}$. But since they are independent, $f_{X_{1}},f_{X_{2}},f_{X_{3}}$ are at the same time marginal probability measures, so the problem becomes finding a joint distribution $g$ of $(X_{1},X_{2},X_{3})$ subject to 3! conditions $$\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}f_{X_{\pi(3)}}(\nu)f_{X_{\pi(2)}}(\omega)f_{X_{\pi(3)}}(\eta)d\nu d\omega d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}g_{X_{\pi(3)},X_{\pi(2)},X_{\pi(1)}}(\nu,\omega,\eta)d\nu d\omega d\eta=p_\pi$$.

If you are willing to assume that $p_{\pi}\equiv\frac{1}{3!}$ then a special case that suffices to work is $X_{i}$ are exchangeable, which deserves an extended discussion like the one in Chap 5,7 in [Kallenberg].

[Kallenberg]Kallenberg, Olav. Probabilistic symmetries and invariance principles. Springer Science & Business Media, 2006.

This is too long to be a comment. But I felt it serves as a pointer.

Without loss of generality we assume $X_{1},X_{2},X_{3}$ share the same image domain $[0,1]$.

For a specific $\pi\in S_{3}$, the probability of $X_{\pi(1)}<X_{\pi(2)}<X_{\pi(3)}$ is $$\int_{0}^{1}dF_{X_{\pi(3)}}(\nu)\int_{0}^{\nu}dF_{X_{\pi(2)}}(\omega)\int_{0}^{\omega}dF_{X_{\pi(1)}}(\eta)=\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta$$where $F_{X_{i}}$ are probability measures on $[0,1]$; $f_{X_{i}}$ are probability densities w.r.t. Lebesgue measure on $[0,1]$, therefore it suffices to make the 3! integral equations have a consistent solution $f_{X_{1}},f_{X_{2}},f_{X_{3}}$. But since they are independent, $f_{X_{1}},f_{X_{2}},f_{X_{3}}$ are at the same time marginal probability measures, so the problem becomes finding a joint distribution $g$ of $(X_{1},X_{2},X_{3})$ subject to 3! conditions $$\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}f_{X_{\pi(3)}}(\nu)f_{X_{\pi(2)}}(\omega)f_{X_{\pi(3)}}(\eta)d\nu d\omega d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}g_{X_{\pi(3)},X_{\pi(2)},X_{\pi(1)}}(\nu,\omega,\eta)d\nu d\omega d\eta$$, which can be done by copula techniques.

If you are willing to assume that $p_{\pi}\equiv\frac{1}{3!}$ then a special case that suffices to work is $X_{i}$ are exchangeable, which deserves an extended discussion like the one in Chap 5,7 in [Kallenberg].

[Kallenberg]Kallenberg, Olav. Probabilistic symmetries and invariance principles. Springer Science & Business Media, 2006.

This is too long to be a comment. But I felt it serves as a pointer.

Without loss of generality we assume $X_{1},X_{2},X_{3}$ share the same image domain $[0,1]$.

For a specific $\pi\in S_{3}$, the probability of $X_{\pi(1)}<X_{\pi(2)}<X_{\pi(3)}$ is $$\int_{0}^{1}dF_{X_{\pi(3)}}(\nu)\int_{0}^{\nu}dF_{X_{\pi(2)}}(\omega)\int_{0}^{\omega}dF_{X_{\pi(1)}}(\eta)=\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta$$where $F_{X_{i}}$ are probability measures on $[0,1]$; $f_{X_{i}}$ are probability densities w.r.t. Lebesgue measure on $[0,1]$, therefore it suffices to make the 3! integral equations have a consistent solution $f_{X_{1}},f_{X_{2}},f_{X_{3}}$. But since they are independent, $f_{X_{1}},f_{X_{2}},f_{X_{3}}$ are at the same time marginal probability measures, so the problem becomes finding a joint distribution $g$ of $(X_{1},X_{2},X_{3})$ subject to 3! conditions $$\int_{0}^{1}f_{X_{\pi(3)}}(\nu)d\nu\int_{0}^{\nu}f_{X_{\pi(2)}}(\omega)d\omega\int_{0}^{\omega}f_{X_{\pi(3)}}(\eta)d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}f_{X_{\pi(3)}}(\nu)f_{X_{\pi(2)}}(\omega)f_{X_{\pi(3)}}(\eta)d\nu d\omega d\eta=\int_{0}^{1}\int_{0}^{\nu}\int_{0}^{\omega}g_{X_{\pi(3)},X_{\pi(2)},X_{\pi(1)}}(\nu,\omega,\eta)d\nu d\omega d\eta$$.

If you are willing to assume that $p_{\pi}\equiv\frac{1}{3!}$ then a special case that suffices to work is $X_{i}$ are exchangeable, which deserves an extended discussion like the one in Chap 5,7 in [Kallenberg].

[Kallenberg]Kallenberg, Olav. Probabilistic symmetries and invariance principles. Springer Science & Business Media, 2006.