I am trying to see "analogy" between the two concepts in the title, although I am familiar with independent definitions of them.
(1) Let $G$ be a group, $N$ a normal subgroup. Define an equivalence relation on $G$ by $g_1\sim g_2$ if $g_1g_2^{-1}\in N$. Let $[g_1]$ denote equivalence class of $g$, and $G/N$ the set of equivalence classes. Then we have a canonical map $\varphi\colon G\rightarrow G/N$, where domain is group, and codomain is "set" right now.
Then the quotient structure on $G/N$ is a "group structure" such that the map $\varphi$ becomes a homomorphism. Such a structure is unique. We call $G/N$ quotient group.
(2) Let $X$ be a topological space, $\sim$ an equivalence relation on $X$, and $[x]$ denote equivalence class. Let $X/{\sim}$ denote the set of equivalence classes. Then there is canonical map $\varphi\colon X\rightarrow X/{\sim}$, where domain is topological space and codomain is just set right now.
The quotient topology on $X/{\sim}$ is a topology such that the map $\varphi$ is continuous.
But, I was feeling that there could be more than one topological structures on $X/\sim$ which make the canonical map continuous. For example, take nice example of quotient topology on one hand, and structure of indiscrete topology on $X/{\sim}$ on the other hand.
Question: What is exact definition of quotient topology which makes it unique in some sense? (compare with quotient structure in (1))
The quotient topology is the unique topology on the set $X/{\sim}$ of equivalence classes such that for any topological space $Y$ and any map $f:X/{\sim}\to Y$, $f$ is continuous iff $\varphi\circ f$ is continuous. (Note that in the case where you take $Y=X/{\sim}$ and $f$ to be the identity map, this says in particular that $\varphi$ itself is continuous.) The analogous characterization also works for the quotient group structure, and pretty much any other quotient structure you'll ever run into.
Another way to phrase this in the case of topological spaces is that the quotient topology is the finest topology that makes $\varphi$ continuous. Indeed, for any topology that makes $\varphi$ continuous, $\varphi\circ f$ will be continuous whenever $f$ is continuous. So assuming $\varphi$ is continuous, the condition above is just that $f$ is continuous whenever $\varphi\circ f$ is: that is, there are as many continuous maps out of $X/{\sim}$ as possible (given the condition that $\varphi$ is continuous). More continuous maps out of a space means a finer topology, so this is saying it's the finest topology for which $\varphi$ is continuous.