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.

Let $S \cong T$ denote the relation that $S$ is (relation) isomorphic to $T$.

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 $S_2$.

Then by Inverse of Relation Isomorphism is Relation Isomorphism, $\phi^{-1}: S_2 \to S_1$ is also a relation isomorphism.

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 Relation Isomorphisms is Relation Isomorphism, the composite mapping $\beta \circ \alpha$ is also a relation isomorphism.

That is, $S_1 \cong S_3$.

So relation isomorphism is transitive.