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 the fact that the Identity Mapping is a Bijection.

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.