Are there infinitely many clones containing all the constant functions?

A clone on a set $A$ is a collection of $m$-ary functions $A^m \to A$ ($m$ depends on the function) that is closed under composition and includes all projection functions. Clones are significant because for any language of possible operations on $A$, the functions definable from those operations is a clone.

Is it true for all $n \in \mathbb{N}$ that there are only finitely many clones containing all constant functions? That is, for each $a \in A$ the clone contains a function $f: A \to A$, such that $f(x) = a$ for all $x$.

Motivation

The classic case is where $A = \{0,1\}$; a set of functions is "expressively adequate" if it generates the clone of all functions. It is a famous and fundamental result that sets such as $\{\lor, \lnot\}$ and $\{\to, \bot\}$ are expressively adequate. A more powerful and incredibly beautiful result is Post's lattice which completely characterizes ALL countably many possible clones on $\{0,1\}$. Wikipedia states: "the lattice of clones on a three-element (or larger) set [has] the cardinality of the continuum, and a complicated inner structure."

But I have an alternate definition of expressive adequacy in mind where we wish to always allow constants $0$ and $1$. Thus we restrict to those clones containing the constant functions, and we find that there are only finitely many clones on $\{0,1\}$. In fact, there are 7 clones:

• $UM$, which contains just constant functions and projections
• $\Lambda$, the set of conjunctive functions
• $V$, the set of disjunctive functions
• $U$, the set of essentially unary functions (depend only one one coordinate)
• $A$, the set of affine functions
• $M$, the set of monotone functions
• $\top$, all functions.

Thus I am interested in whether there are still finitely many such clones for a set of size greater than $2$.