A variety of algebras satisfying some dual conditions I would like prove that, under the conditions described below, no non-trivial variety exists.

Let $\mathcal{V}$ be a variety of algebras e.g. rings, semigroups, semilattices.

Further suppose that:

  
*
  
*The empty algebra exists i.e. $\mathcal{V}$ has no constants.
  
*The dual condition also holds. That is, let $\mathtt{1} \in \mathcal{V}$ be the one element terminal algebra. Then $id_\mathtt{1}$ is the only homomorphism whose domain is $\mathtt{1}$.
  

Any help much appreciated.
 A: It's false.  Take the theory generated by unary operations $\zeta$ and $\iota$ and a binary operation $\cdot$, subject to the equations
$$
\zeta(x) = \zeta(y),
\qquad
\iota(x) = \iota(y),
\qquad
\zeta(x) \cdot y = \zeta(x),
\qquad
\iota(x) \cdot y = y.
$$
An example of an algebra is any ring, with $\zeta(x) = 0$, $\iota(x) = 1$, and the usual $\cdot$.  So the theory is not trivial.
Obviously $\emptyset$ is an algebra.  Now suppose we have a homomorphism from $1$ to an algebra $A$, picking out an element $a \in A$.  The equation $\iota(x) = \zeta(x)$ holds in $1$, so $\iota(a) = \zeta(a)$ too.  Hence for all $b \in A$,
$$
b = \iota(a) \cdot b = \zeta(a) \cdot b = \zeta(a),
$$
giving $A = \{\zeta(a)\} \cong 1$.  
The trick is that although the theory has no actual constants, it has two "pseudo-constants", namely, $\zeta$ and $\iota$.
A: For varieties that only have unary function symbols, it is necessary to use constant functions as in Tom Leinster's answer.
$\mathbf{Proposition}$ Suppose that $V$ is a variety such that all the function symbols are unary and for all terms $t$, $V$ does not satisfy the identity $t(x)=t(y)$. Then there is some algebra $X\in V$ and multiple homomorphisms $1\rightarrow X$.
$\mathbf{Proof}$ Suppose that $V$ does not satisfy the identity $t(x)=t(y)$ for each term $t$. Then there does not exist terms $s,t$ that satisfy the identity $s(x)=t(y)$. Therefore every identity in $V$ is an identity of the form $s(x)=t(x)$. Let $G$ be the monoid of all terms in $V$ in one variable up to logical equivalence. Then $G$ is a monoid defined by $s(x)t(x)=s(t(x))$. We may therefore think of the variety $V$ as the collection of all actions from the monoid $G$ to some set $X$. On the other hand, let $G$ be the trivial action on a set $X$. In other words, $gx=x$ for all $g\in G$. Then every function $\phi:1\rightarrow G$ is a homomorphism.
