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I think it is not possible. Such a ring isomorphism $\Phi$ should also preserve the order structure, because in both rings non-negative elements are exactly the squares; as a consequence, it must also preserve the constant functions, since it preserves the constant $1$. In other words, $\Phi$ is an ordered $\mathbb{R}$-algebras isomorphism.

In both rings, characteristic functions of singletons can be characterized in terms of the ordered $\mathbb{R}$-algebra structure, as e.g. those idempotents $u$ such that any positive element smaller than $u$ is a scalar multiple of $u$ ( that is "$0\le v\le u$ implies $v=\lambda u$ ").

Note that the ring $S$ contains all characteristic functions of singletons of $\mathbb{R}$.

Since $\Phi$ preserves the ordered $\mathbb{R}$-algebras structure, if $u:=\chi_{\{t\}}$ is a characteristic function of a singleton of $\mathbb{R}$, then $\Phi(u)$ is also a characteristic function of a singleton $\chi_{\{x\}}$ of $X$.

This way we have defined an injective map $\phi:\mathbb{R}\to X$ such that for all $t\in\mathbb{R}$ one has $\Phi(\chi_{\{t\}})= \chi_{\{\phi(t)\}}\, .$ Note that by the order properties of $\Phi$, if $f\in S$ vanishes at $t_0\in\mathbb{R}$, then $\Phi(f)$ vanishes at $\phi(t_0)\in X$ (reason: if $f(t_0)=0$ then $f ^2\ge\lambda\chi_{\{t_0\}}$ for no $\lambda>0$, so $\Phi(f )^2\ge\Phi(\lambda\chi_{\{t_0\}})=\lambda\chi_{\{\phi(t_0)\}}$ for no $\lambda>0$, hence $\Phi(f)(\phi(t_0))=0$). Since $\Phi(c)=c$ for any constant function, we also have $\Phi(f(\phi(r)))=f(r)$ for any $f\in S$ and $r\in\mathbb{R}$ (reason: if $c:=f(r)$, the function $f-c$ vanishes in the point $r$, so that $\Phi( f -c) = \Phi(f) - c$ vanishes in $\phi(r)$, that is $\Phi( f)(\phi(r)) = f(r)$ for all $f\in S$ and $r\in \mathbb{R}$. So $\Phi^{-1}(u)=u\circ\phi$ for any $u\in C(X)$ (as $\Phi$ is bijective). However this yields a contradiction.

Let $\{q_n\}_{n\in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$. Then the (normally convergent) series $\sum_{n\in\mathbb{N}} 2^{-n} \chi_{\psi(q_n)}$ represents an element $u$ of $C(X)$ that for any $t\in\mathbb{R}$ vanishes at $\phi(t)$ if and only in $t$ is irrational; hence $u\circ \phi\in S$ vanishes exactly on the irrationals, a contradiction.

I think it is not possible. Such a ring isomorphism $\Phi$ should also preserve the order structure, because in both rings non-negative elements are exactly the squares; as a consequence, it must also preserve the constant functions, since it preserves the constant $1$. In other words, $\Phi$ is an ordered $\mathbb{R}$-algebras isomorphism.

In both rings, characteristic functions of singletons can be characterized in terms of the ordered $\mathbb{R}$-algebra structure, as e.g. those idempotents $u$ such that any positive element smaller than $u$ is a scalar multiple of $u$ ( that is "$0\le v\le u$ implies $v=\lambda u$ ").

Note that the ring $S$ contains all characteristic functions of singletons of $\mathbb{R}$.

Since $\Phi$ preserves the ordered $\mathbb{R}$-algebras structure, if $u:=\chi_{\{t\}}$ is a characteristic function of a singleton of $\mathbb{R}$, then $\Phi(u)$ is also a characteristic function of a singleton $\chi_{\{x\}}$ of $X$.

This way we have defined an injective map $\phi:\mathbb{R}\to X$ such that for all $t\in\mathbb{R}$ one has $\Phi(\chi_{\{t\}})= \chi_{\{\phi(t)\}}\, .$ Note that by the order properties of $\Phi$, if $f\in S$ vanishes at $t_0\in\mathbb{R}$, then $\Phi(f)$ vanishes at $\phi(t_0)\in X$ (reason: if $f(t_0)=0$ then $f ^2\ge\lambda\chi_{\{t_0\}}$ for no $\lambda>0$, so $\Phi(f )^2\ge\Phi(\lambda\chi_{\{t_0\}})=\lambda\chi_{\{\phi(t_0)\}}$ for no $\lambda>0$, hence $\Phi(f)(\phi(t_0))=0$). Since $\Phi(c)=c$ for any constant function, we also have $\Phi(f(\phi(r)))=f(r)$ for any $f\in S$ and $r\in\mathbb{R}$ (reason: if $c:=f(r)$, the function $f-c$ vanishes in the point $r$, so that $\Phi( f -c) = \Phi(f) - c$ vanishes in $\phi(r)$, that is $\Phi( f)(\phi(r)) = f(r)$ for all $f\in S$ and $r\in \mathbb{R}$). So $\Phi^{-1}(u)=u\circ\phi$ for any $u\in C(X)$ (as $\Phi$ is bijective). However this yields a contradiction.

Let $\{q_n\}_{n\in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$. Then the (normally convergent) series $\sum_{n\in\mathbb{N}} 2^{-n} \chi_{\psi(q_n)}$ represents an element $u$ of $C(X)$ that for any $t\in\mathbb{R}$ vanishes at $\phi(t)$ if and only in $t$ is irrational; hence $u\circ \phi\in S$ vanishes exactly on the irrationals, a contradiction.

I think it is not possible. Such a ring isomorphism $\Phi$ should also preserve the order structure, because in both rings non-negative elements are exactly the squares; as a consequence, it must also preserve the constant functions, since it preserves the constant $1$. In other words, $\Phi$ is an ordered $\mathbb{R}$-algebras isomorphism.

In both rings, characteristic functions of singletons can be characterized in terms of the ordered $\mathbb{R}$-algebra structure, as e.g. those idempotents $u$ such that any positive element smaller than $u$ is a scalar multiple of $u$ ( that is "$0\le v\le u$ implies $v=\lambda u$ ").

Note that the ring $S$ contains all characteristic functions of singletons of $\mathbb{R}$.

Since $\Phi$ preserves the ordered $\mathbb{R}$-algebras structure, if $u:=\chi_{\{t\}}$ is a characteristic function of a singleton of $\mathbb{R}$, then $\Phi(u)$ is also a characteristic function of a singleton $\chi_{\{x\}}$ of $X$.

This way we have defined an injective map $\phi:\mathbb{R}\to X$ such that for all $t\in\mathbb{R}$ one has $\Phi(\chi_{\{t\}})= \chi_{\{\phi(t)\}}\, .$ Note that by the order properties of $\Phi$, if $f\in S$ vanishes at $t_0\in\mathbb{R}$, then $\Phi(f)$ vanishes at $\phi(t_0)\in X$ (reason: if $f(t_0)=0$ then $f ^2\ge\lambda\chi_{\{t_0\}}$ for no $\lambda>0$, so $\Phi(f )^2\ge\Phi(\lambda\chi_{\{t_0\}})=\lambda\chi_{\{\phi(t_0)\}}$ for no $\lambda>0$, hence $\Phi(f)(t_0)=0$). Since $\Phi(c)=c$ for any constant function, we also have $\Phi(f(\phi(r)))=f(r)$ for any $f\in S$ and $r\in\mathbb{R}$ (reason: if $c:=f(r)$, the function $f-c$ vanishes in the point $r$, so that $\Phi( f -c) = \Phi(f) - c$ vanishes in $\phi(r)$, that is $\Phi( f)(\phi(r)) = f(r)$. So $\Phi^{-1}(u)=u\circ\phi$ for any $u\in C(X)$. However this yields a contradiction.

Let $\{q_n\}_{n\in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$. Then the (normally convergent) series $\sum_{n\in\mathbb{N}} 2^{-n} \chi_{\psi(q_n)}$ represents an element $u$ of $C(X)$ that for any $t\in\mathbb{R}$ vanishes at $\phi(t)$ if and only in $t$ is irrational; hence $u\circ \phi\in S$ vanishes exactly on the irrationals, a contradiction.

I think it is not possible. Such a ring isomorphism $\Phi$ should also preserve the order structure, because in both rings non-negative elements are exactly the squares; as a consequence, it must also preserve the constant functions, since it preserves the constant $1$. In other words, $\Phi$ is an ordered $\mathbb{R}$-algebras isomorphism.

In both rings, characteristic functions of singletons can be characterized in terms of the ordered $\mathbb{R}$-algebra structure, as e.g. those idempotents $u$ such that any positive element smaller than $u$ is a scalar multiple of $u$ ( that is "$0\le v\le u$ implies $v=\lambda u$ ").

Note that the ring $S$ contains all characteristic functions of singletons of $\mathbb{R}$.

Since $\Phi$ preserves the ordered $\mathbb{R}$-algebras structure, if $u:=\chi_{\{t\}}$ is a characteristic function of a singleton of $\mathbb{R}$, then $\Phi(u)$ is also a characteristic function of a singleton $\chi_{\{x\}}$ of $X$.

This way we have defined an injective map $\phi:\mathbb{R}\to X$ such that for all $t\in\mathbb{R}$ one has $\Phi(\chi_{\{t\}})= \chi_{\{\phi(t)\}}\, .$ Note that by the order properties of $\Phi$, if $f\in S$ vanishes at $t_0\in\mathbb{R}$, then $\Phi(f)$ vanishes at $\phi(t_0)\in X$ (reason: if $f(t_0)=0$ then $f ^2\ge\lambda\chi_{\{t_0\}}$ for no $\lambda>0$, so $\Phi(f )^2\ge\Phi(\lambda\chi_{\{t_0\}})=\lambda\chi_{\{\phi(t_0)\}}$ for no $\lambda>0$, hence $\Phi(f)(\phi(t_0))=0$). Since $\Phi(c)=c$ for any constant function, we also have $\Phi(f(\phi(r)))=f(r)$ for any $f\in S$ and $r\in\mathbb{R}$ (reason: if $c:=f(r)$, the function $f-c$ vanishes in the point $r$, so that $\Phi( f -c) = \Phi(f) - c$ vanishes in $\phi(r)$, that is $\Phi( f)(\phi(r)) = f(r)$ for all $f\in S$ and $r\in \mathbb{R}$. So $\Phi^{-1}(u)=u\circ\phi$ for any $u\in C(X)$ (as $\Phi$ is bijective). However this yields a contradiction.

Let $\{q_n\}_{n\in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$. Then the (normally convergent) series $\sum_{n\in\mathbb{N}} 2^{-n} \chi_{\psi(q_n)}$ represents an element $u$ of $C(X)$ that for any $t\in\mathbb{R}$ vanishes at $\phi(t)$ if and only in $t$ is irrational; hence $u\circ \phi\in S$ vanishes exactly on the irrationals, a contradiction.

I think it is not possible. Such a ring isomorphism $\Phi$ should also preserve the order structure, because in both rings non-negative elements are exactly the squares; as a consequence, it must also preserve the constant functions, since it preserves the constant $1$. In other words, $\Phi$ is an ordered $\mathbb{R}$-algebras isomorphism.

In both rings, characteristic functions of singletons can be characterized in terms of the ordered $\mathbb{R}$-algebra structure, as e.g. those idempotents $u$ such that any positive element smaller than $u$ is a scalar multiple of $u$ ( that is "$0\le v\le u$ implies $v=\lambda u$ ").

Note that the ring $S$ contains all characteristic functions of singletons of $\mathbb{R}$.

Since $\Phi$ preserves the ordered $\mathbb{R}$-algebras structure, if $u:=\chi_{\{t\}}$ is a characteristic function of a singleton of $\mathbb{R}$, then $\Phi(u)$ is also a characteristic function of a singleton $\chi_{\{x\}}$ of $X$.

This way we have defined an injective map $\phi:\mathbb{R}\to X$ such that for all $t\in\mathbb{R}$ one has $\Phi(\chi_{\{t\}})= \chi_{\{\phi(t)\}}\, .$ Note that by the order properties of $\Phi$, if $f\in S$ vanishes at $t_0\in\mathbb{R}$, then $\Phi(f)$ vanishes at $\phi(t_0)\in X$ (reason: if $f(t_0)=0$ then $f ^2\ge\lambda\chi_{\{t_0\}}$ for no $\lambda>0$, so $\Phi(f )^2\ge\Phi(\lambda\chi_{\{t_0\}})=\lambda\chi_{\{\phi(t_0)\}}$ for no $\lambda>0$, hence $\Phi(f)(t_0)=0$). Since $\Phi(c)=c$ for any constant function, we also have $\Phi(f(\phi(t)))=f(t)$ for any $f\in S$ and $t\in\mathbb{R}$, that is $\Phi^{-1}(u)=u\circ\phi$ for any $u\in C(X)$. However this yields a contradiction.

Let $\{q_n\}_{n\in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$. Then the (normally convergent) series $\sum_{n\in\mathbb{N}} 2^{-n} \chi_{\psi(q_n)}$ represents an element $u$ of $C(X)$ that for any $t\in\mathbb{R}$ vanishes at $\phi(t)$ if and only in $t$ is irrational; hence $u\circ \phi\in S$ vanishes exactly on the irrationals, a contradiction.

I think it is not possible. Such a ring isomorphism $\Phi$ should also preserve the order structure, because in both rings non-negative elements are exactly the squares; as a consequence, it must also preserve the constant functions, since it preserves the constant $1$. In other words, $\Phi$ is an ordered $\mathbb{R}$-algebras isomorphism.

In both rings, characteristic functions of singletons can be characterized in terms of the ordered $\mathbb{R}$-algebra structure, as e.g. those idempotents $u$ such that any positive element smaller than $u$ is a scalar multiple of $u$ ( that is "$0\le v\le u$ implies $v=\lambda u$ ").

Note that the ring $S$ contains all characteristic functions of singletons of $\mathbb{R}$.

Since $\Phi$ preserves the ordered $\mathbb{R}$-algebras structure, if $u:=\chi_{\{t\}}$ is a characteristic function of a singleton of $\mathbb{R}$, then $\Phi(u)$ is also a characteristic function of a singleton $\chi_{\{x\}}$ of $X$.

This way we have defined an injective map $\phi:\mathbb{R}\to X$ such that for all $t\in\mathbb{R}$ one has $\Phi(\chi_{\{t\}})= \chi_{\{\phi(t)\}}\, .$ Note that by the order properties of $\Phi$, if $f\in S$ vanishes at $t_0\in\mathbb{R}$, then $\Phi(f)$ vanishes at $\phi(t_0)\in X$ (reason: if $f(t_0)=0$ then $f ^2\ge\lambda\chi_{\{t_0\}}$ for no $\lambda>0$, so $\Phi(f )^2\ge\Phi(\lambda\chi_{\{t_0\}})=\lambda\chi_{\{\phi(t_0)\}}$ for no $\lambda>0$, hence $\Phi(f)(t_0)=0$). Since $\Phi(c)=c$ for any constant function, we also have $\Phi(f(\phi(r)))=f(r)$ for any $f\in S$ and $r\in\mathbb{R}$ (reason: if $c:=f(r)$, the function $f-c$ vanishes in the point $r$, so that $\Phi( f -c) = \Phi(f) - c$ vanishes in $\phi(r)$, that is $\Phi( f)(\phi(r)) = f(r)$. So $\Phi^{-1}(u)=u\circ\phi$ for any $u\in C(X)$. However this yields a contradiction.

Let $\{q_n\}_{n\in \mathbb{N}}$ be an enumeration of $\mathbb{Q}$. Then the (normally convergent) series $\sum_{n\in\mathbb{N}} 2^{-n} \chi_{\psi(q_n)}$ represents an element $u$ of $C(X)$ that for any $t\in\mathbb{R}$ vanishes at $\phi(t)$ if and only in $t$ is irrational; hence $u\circ \phi\in S$ vanishes exactly on the irrationals, a contradiction.