Under what circumstances $\langle g^m \rangle = \langle g^n \rangle$ implies $g^m = g^n$? 
Let $G$ be a group, and $g\in G$. Assume that $$\langle g^m \rangle = \langle g^n \rangle$$ for some $m, n \in \mathbb{Z}$.

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*When can we deduce that $g^m = g^n$?


*What if $G$ is cyclic? Finite? Cyclic and finite?

For an aribtrary group $G$ and an element $g$ of a finite order $|g|$ we have:
$\langle g^m \rangle = \langle g^n \rangle \Rightarrow g^m \in \langle g^n \rangle$ and $g^n \in \langle g^m \rangle \Rightarrow \exists q, r \in \mathbb{Z}: \ \ n \equiv qm \mod |g|$ and $rn \equiv m \mod |g|$.
If $|g| = \infty$, then $m|n$ and $n|m$, and $m = n$, so, obviously, $g^m = g^n$.
But what if $|g| \in \mathbb{N}$?

I take the question to be:

What is the condition on $g$ so that the following holds:
  For all $m,n\in\mathbb Z$ we have $\langle g^m\rangle = \langle g^n\rangle \implies g^m = g^n$.

Given that $\langle g\rangle = \langle g^{-1}\rangle$, this obviously implies that $g=g^{-1}$.On the other hand, if $g=g^{-1}$, then $g^n\in\{e,g\}$, and clearly, $\langle e\rangle=\langle g\rangle$ if and only if $e=g$. Therefore the reverse direction also holds, and therefore the answer is:

The condition holds if and only if $g=g^{-1}$. That is, $g$ is either the neutral element, or an element of order $2$.

