Necessary conditions of finite primal algebra

Question

An algebra $\mathbf A$ is called primal if every function of any arity on $A$ is a term function (in short $T^{\mathbf A} = F(A)$).

Let $\mathbf A$ be finite and primal. Then $\mathbf A$ is its only subalgebra, $\mathbf A$ has no non-trivial congruences (is simple) and the only automorphism on $\mathbf A$ is the identity.

The first two are clear. Assume $|A|\geqslant2$

1. Suppose $\mathbf B\leqslant A$ such that $B\subset A$. Pick any map $f:A\to A$ such that $f(B)\not\subseteq B$. Primality implies $f\in T_1^{\mathbf A}$, but then $f\vert _B \in T_1^{\mathbf B}$, which implies $f(B)\subseteq B$.
2. Suppose $\rho\in\mathrm{Con}(\mathbf A)$ and assume there exist $a,b,c,d\in A$ such that $(c,d)\notin \rho$ and $(a,b)\in\rho$, $a\neq b$. Pick any $f:A\to A$ such that $f(a) = c$ and $f(b) = d$. Since term functions are compatible with congruences, we have $(c,d)\in\rho$.

But how does the third item follow? I'm possibly missing something obvious here.

Answer 1

Let $f$ be an automorphism on $\mathbf A$. Assume for a contradiction $f(a)=b$ for some $a\neq b$. The constant function $c_b$ is a term function. Hence $$ f(a) = b = c_{b}(f(a)) = f(c_b(a)) = f(f(a)) $$ which implies $a=b$.