Fully invariant congruence

Remarks

The congruences $\Delta_A$ and $\nabla_A$ are always fully invariant, and so is the intersection of any family of fully invariant congruences.

Example: on the group $\Z^2 = \Z \times \Z$, congruence modulo $2$ in both coordinates is fully invariant. Every endomorphism of $\Z^2$ is given by an integer matrix $M$, and $a - b \in 2\Z^2$ implies $M a - M b = M(a - b) \in 2\Z^2$.

Non-example: on $\Z^2$, let $(a, b) \in \theta$ when the first coordinates of $a$ and $b$ are equal. This is a congruence, but the swap $(x, y) \mapsto (y, x)$ is an endomorphism that sends the related pair $((0, 0), (0, 1))$ to the unrelated pair $((0, 0), (1, 0))$.

On the term algebra $\mathbf{T}(X)$ over a set of variables $X$, the endomorphisms are the substitutions. So the fully invariant congruences on $\mathbf{T}(X)$ are the equational theories in the variables $X$.