$\langle a\rangle \simeq \langle b \rangle$ if and only if $a$ and $b$ have the same index and period 
If $a$ and $b$ are elements of finite order in the same or in different semigroups, the  $\langle a\rangle  \simeq \langle b \rangle$ if and only if $a$ and $b$ have the same index and period.

$\Leftarrow]$ we have: $\langle a\rangle=\{a, a^2, \dots, a^m, a^{m+1}, \dots, a^{m+r-1}\}$ and $\langle b\rangle =\{b, b^2, \dots , b^m, b^{m+1}, \dots, b^{m+r-1}\}$ where $m$ is the index and $r$ is the period of $a$ and $b$, which by hypothesis is the same. Let's consider the function
$f: \langle a \rangle \to \langle b \rangle$ by $f(a^{i})= b^{i}$ it is easy to verify that $f$ is an isomorphism.
I have trouble showing that if $\langle a \rangle$ and $\langle b \rangle$ are isomorphic then  the index and period of $ a$ are equal to the indexes and period of $b$.
I think that if we assume that they are not true, it may contradict the minimality of the period and the index, but I do not know if it is true, any suggestions? thanks
 A: Let $a$ have period $r$ and index $m$; let $b$ have period $s$ and index $n$. We want to show that $r=s$ and $m=n$.
Note that if $a^i=a^j$ for some $i\neq j$, then it must be the case that $i\equiv j\pmod{r}$ and $i,j\geq m$; symmetrically, if $b^u=b^w$ with $u\neq w$, then $u\equiv w\pmod {s}$ and $u,w\geq n$.
Let $f\colon \langle a\rangle \to \langle b\rangle$ be an isomorphism, and let $g$ be the inverse of $f$.
Say $f(a)=b^k$ and $g(b)=a^h$. Then
$$a=g(f(a)) = g(b^k) = (g(a))^k = a^{hk}.$$
If either $h$ or $k$ are greater than $1$, then $hk\neq 1$. Therefore, $1\geq m$ (so $m=1$), and $hk\equiv 1\pmod{r}$; symmetrically, since
$b=f(g(b)) = f(a^h) = f(a)^h = b^{kh}$, then $1\geq n$ (so $n=1$) and $kh\equiv 1\pmod{s}$. Thus we have $m=n=1$. But then $|\langle a\rangle| = r$, $|\langle b\rangle|=s$, and since the two are isomorphic, then $r=s$.
Thus we may assume that $f(a)=b$ and $g(b)=a$. Can you finish it in this case?
(Note: we need to treat the case $m=1$ separately from $m\neq 1$ (symmetrically for $n$), because whenb $m=1$ we have a group, and $\langle a\rangle$ may have more than one generator; it is a cyclic group, after all. But when $m\gt 1$ the automorphisms must send $a$ to itself, as that is the only generator.)
