Let $A\_n = \{(1,\ldots,n) , f \}$ where $f(i) = (i+1)$ if $i \neq n $ otherwise $f(n) = 1$.
This describes a mono unary algebra.
The proof for $HPS \neq SPHS$ I know uses metabelian groups and was published by 
George Bergman in Algebra Universalis. 
Here I need to prove $SHPS \neq SPHS$ for the class of mono unary algebra $R$ = { $A\_p$ | where $p$ is a prime.}


I do not seem to make a start using the above algebra. 



SHPS and SPHS inequality using monounary algebra
The only subalgebras of $A\_p$ are $\varnothing$ and $A\_p$. I claim that the only homomorphic images of $A\_p$ are $A\_p$ and the one-element algebra.


**Note:** there was an arithmetical error in the following paragraph. Fixed.


Indeed, let $\Phi$ be a congruence on $A\_p$ that contains a pair $(a,b)$ with $a\neq b$. Applying $f$ sufficient times, we get a pair $(1,k)\in\Phi$, $1\lt k\leq p$. If $k=p$, then applying $f$ we get $(2,1)$ so $(1,2)\in\Phi$, and so after applying $f$ we again we obtain $(i,i+1)\in\Phi$ for all $i$, and since $\Phi$ is an equivalence relation, it is the total relation. If $2\leq k \lt p$, then there exists $n$, $2\leq n\lt p$, such that $nk\equiv 1\pmod{p}$. Applying $f$ $(n-1)k$ times to $(1,k)$, we obtain a pair whose second entry is $1$, and whose first entry is congruent to $(n-1)k$ modulo $p$. Since $nk\equiv 1\pmod{p}$, then $(n-1)k\equiv 1-k\equiv p+1-k\pmod{p}$, so we have shown that the pair $(p+1-k,1)\in\Phi$. On the other hand, applying $f$ $p-k+1$ times to $(1,k)$ we obtain $(p-k+2,1)\in\Phi$, so we get $(p-k+1,p-k+2)\in\Phi$ and so we must again have that $\Phi$ is the total congruence.


Therefore, $H(R) = R\cup\{A\_1\}$. Since a product that contains copies of $A\_1$ is isomorphic to either $A\_1$ or one which contains no copies of $A\_1$, we conclude that $PH(R) = P(R)\cup\{A\_1\}$. So $SPHS(R) = SP(R)\cup\{A\_1\}$. The subalgebras are isomorphic to disjoint unions of (an arbitrary number of) copies of $A\_k$s and $\mathbb{N}$, where $k$ is squarefree. Indeed, if $k=p\_1\cdots p\_r$ is a product of distinct primes, then by taking $A\_{p\_1}\times\cdots\times A\_{p\_r}$ and considering the subalgebra generated by $(1,\ldots,1)$, we obtain an element of $f$-order $k$, and so we get an algebra isomorphic to $A\_k$; hence, all such $A\_k$ are subalgebras of products of elements of $R$. Also, by taking the product $\prod A\_p$ over all primes and the subalgebra generated by $(1,1,\ldots)$ we obtain an element of infinite $f$-order, which is essentially the same as $(\mathbb{N},s)$. Taking the subalgebra generated by a finite collecion of such generators in distinct components gives the disjoint union. So all described algebras are in $SP(R)$. Conversely, given a subalgebra of a product of elements of $R$, we can partition it into $f$-orbits; an element $(a\_i)$ has $f$-order $n$ if and only if the lcm of the $f$-orders of the $a\_i$ is $n$, so the finite $f$-orbits of $R$ must be of squarefree order (since those are the possible orders of the components); and infinite orbits give copies of $(\mathbb{N},s)$. 


So $SPHS(R)$ consists of disjoint unions (possibly empty) of copies of $A\_k$s for $k$ squarefree (including $k=1$), and $(\mathbb{N},s)$.


However, $PS(R)$ contains $A=\prod A\_p$, with the product over all primes, and I claim that a homomorphic image of this algebra is the disjoint union of $A\_4$ and $A\_1$, proving that $SHPS(R)\neq SPHS(R)$.


Indeed, partitioning $A$ into disjoint $f$-orbits, define the congruence $\Phi$ which restricted to the subalgebra generated by $(1,1,\ldots)$ (which is isomorphic to $\mathbb{N}$) yields the congruence such that $\mathbb{N}/\Psi\cong A\_4$ (identifying $a\sim b$ if and only if $|a-b|$ is a multiple of $4$), and that identifies all other elements of all other $f$-orbits into a single point. Then $A/\Phi\cong A\_4\coprod A\_1$.


