Relation Isomorphism is Equivalence Relation

Theorem
Relation isomorphism is an equivalence relation.

Proof
Let $\left({S_1, \mathcal R_1}\right)$, $\left({S_2, \mathcal R_2}\right)$ and $\left({S_3, \mathcal R_3}\right)$ be relational structures.

Checking in turn each of the criteria for equivalence:

Reflexive
The fact that relation isomorphism is reflexive follows immediately from Identity Mapping is Relation Isomorphism.

Symmetric
Suppose $S_1 \cong S_2$.

Let $\phi: S_1 \to S_2$ be a relation isomorphism from $S_1$ to $S_2$.

By definition, $\phi$ is a bijection.

Then by Bijection iff Inverse is Bijection, $\phi^{-1}: S_2 \to S_1$ is also a bijection.

Suppose $s_2, t_2 \in \left({S_2, \mathcal R_2}\right)$ such that $\phi^{-1} \left({s_2}\right) = s_1, \phi^{-1} \left({t_2}\right) = t_1$ where $s_1, t_1 \in \left({S_1, \mathcal R_1}\right)$.

It follows that $\left({s_2, t_2}\right) \in \mathcal R_2$ iff $\left({\phi \left({t_1}\right), \phi \left({t_1}\right)}\right) \in \mathcal R_2$.

Since $S_1 \cong S_2$, it follows that $\left({\phi \left({s_1}\right), \phi \left({t_1}\right)}\right) \in \mathcal R_2$ iff $\left({\phi^{-1} \left({s_2}\right), \phi^{-1} \left({t_2}\right)}\right) \in \mathcal R_1$.

That is, $\left({\phi \left({t_1}\right), \phi \left({t_1}\right)}\right) \in \mathcal R_2$ iff $\left({s_1, t_1}\right) \in \mathcal R_1$.

So $\left({s_2, t_2}\right) \in \mathcal R_2$ iff $\left({\phi^{-1} \left({s_2}\right), \phi^{-1} \left({t_2}\right)}\right) \in \mathcal R_1$.

So $S_2 \cong S_1$ and so relation isomorphism is symmetric.

Transitive
Suppose $S_1 \cong S_2$ and $S_2 \cong S_3$.

Let:
 * $\alpha: S_1 \to S_2$ be a relation isomorphism from $S_1$ to $S_2$;
 * $\beta: S_2 \to S_3$ be a relation isomorphism from $S_2$ to $S_3$.

From Composite of Bijections, the composite mapping $\beta \circ \alpha$ is also a bijection.

Let $\left({s_1, t_1}\right) \in \mathcal R_1$.

Suppose:
 * $\alpha \left({s_1}\right) = s_2, \alpha \left({t_1}\right) = t_2$;
 * $\beta \left({s_2}\right) = s_3, \beta \left({t_2}\right) = t_3$.

Since $\alpha$ and $\beta$ are isomorphisms:
 * $\left({s_1, t_1}\right) \in \mathcal R_1$ iff $\left({\alpha \left({s_1}\right), \alpha \left({t_1}\right)}\right) = \left({s_2, t_2}\right) \in \mathcal R_2$;
 * $\left({s_2, t_2}\right) \in \mathcal R_2$ iff $\left({\beta \left({s_2}\right), \beta \left({t_2}\right)}\right) = \left({s_3, t_3}\right) \in \mathcal R_3$.

Thus $\left({s_1, t_1}\right) \in \mathcal R_1$ iff $\left({\left({\beta \circ \alpha}\right) \left({s_1}\right) = \left({\beta \circ \alpha}\right) \left({t_1}\right)}\right) = \left({s_3, t_3}\right) \in \mathcal R_3$.

Hence $\beta \circ \alpha$ is an isomorphism, and so $S_1 \cong S_3$.

So relation isomorphism is transitive.