Mono unary algebra and their product 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.
Now I wish to see the structure of $A_3\times A_5$.
Clearly  $A_3\times A_5 = \{(a,b) : a  \in A_3\text{ and }b \in A_5\}$. Similarly $f^{A_3 \times A_5}$ is $\langle f^{A_3},f^{A_5}\rangle$. But how to visualize the structure of this new algebra ?
How does it differ from another algebra say $A_6\times A_{15}$.
Any help with visualization will be greatly appreciated. 

The structure of $A_3 \times A_5$ is just that of $A_{15}$, as you can see in the picture.
Here I'm using the graphical description of a mono-unary algebra, where the nodes are the elements of the algebra and $u \rightarrow v$ iff $f(u) = v$.

It is just a matter of unfolding the graph.
I didn't add the labels in the product, but I think it is clear...
More interesting can be to give a general description of products of such algebras.

Let $\mathbf{A}_i = \langle A_i, f \rangle$ and $\mathbf{A}_j = \langle A_j, f\rangle$ be two such algebras, where $f(x) = x+1 \mod i$ or $\mod j$ accordingly.
Then $\mathbf{A}_i \vDash f^{i}(x) \approx x$, and $\mathbf{A}_j \vDash f^{j}(x) \approx x$.
Let $d = \gcd(i,j)$ and $m = \mathrm{lcm}(i,j)$.
It follows that
$$\mathbf{A} = \mathbf{A}_i \times \mathbf{A}_j \vDash f^m(x) \approx x.$$
So $\mathbf{A}$ is composed by cycles, each of which has an order that divides $m$, that is, $\mathbf{A}$ has different connected components (each cycle), $B$, and $|B|$ divides $m$.
On the other hand, both $\mathbf{A}_i$ and $\mathbf{A}_j$ are homomorphic images of $\mathbf{A}$.
Since $\mathbf{A}_i$ and $\mathbf{A}_j$ are connected and $\mathbf{A}$ is made of, at least, $d$ connected components, we need congruences $\alpha_i, \alpha_j$ on $\mathbf{A}$ such that 
$$\mathbf{A}/\alpha_i \cong \mathbf{A}_i \quad\text{ and }\quad 
\mathbf{A}/\alpha_j \cong \mathbf{A}_j.$$  
But then, for each connected component $B$ of $\mathbf{A}$, we have 
$$\mathbf{B}/\alpha_i \cong \mathbf{A}_i \quad\text{ and }\quad 
\mathbf{B}/\alpha_j \cong \mathbf{A}_j.$$
It is immediate that in such algebras, for each congruence, all the congruence classes have the same number of elements, and so $|B|$ is a multiple of both $|A_i|$ and $|A_j|$, so that $|B|$ is a multiple of $m$.
Thus, $|B|=m$ and $\mathbf{A}$ is made of $d$ disjoint copies of $\mathbf{A}_m$.

So, generally, for $i, j \geq 1$, if $d = \gcd(i,j)$ and $m = \mathrm{lcm}(i,j)$, then
$$\mathbf{A}_i \times \mathbf{A}_j \cong d \mathbf{A}_m,$$
where a natural number $n$ multiplied by (the graph of) an algebra means $n$ disjoint copies of that algebra.
One possibly interesting consequence of this is that
$$\mathbf{A}_n^k \cong n^{k-1} \mathbf{A}_n.$$
