I will call the subgroup $H$ because the condition that it be normal in $G$ is not important. In fact it could just be a subset $Y$ which is large enough. In the following examples I will use $G=S_{\mathbb{N}},$ the uncountable group of permutations of the positive integers. The action(s) defined will also satisfy the  stronger condition that for any $g_1 \neq g_2$ in $G$ (rather than $g_1,g_2$ in a common coset of $H$) there is an $h$ with $g_1 \cdot h \neq g_2\cdot h.$

My first example will use a normal subgroup and a familiar action. Let $H$ be the subgroup of permutations which fix all but finitely many integers (which is countable and normal). Note the important (to this construction) property that the center of $H$ is trivial. Then the action $g \cdot h=ghg^{-1}$  is a natural group action of $G$ on $H.$ What you desire is that $a$ and $ah_1$ never act in the same way on $H$ when $h_1$ is not the identity. Pick an $h_2 \in H$ such that $h_1h_2 \neq h_2h_1$ and observe that $a\cdot h_2 \neq (ah_1) \cdot h_2.$ This construction works any time you have a group $G$ with a normal subgroup $H$ whose center is trivial. In this particular case the stronger property condition is satisfied because the center of the entire group is trivial.

The next construction is quite arbitrary and the action not that natural. I will give it in generality and then get more specific. I need some infinite subgroup $H$ along with a countable subset $Y$ (which could be all of $H$ but need not be.) At this point only $Y$ matters, the rest of $H$ doesn't. I do require that there be an enumeration $y_1,y_2,\cdots$ of the elements of $Y.$ The enumeration should be constructive, but can be as simple or complex as you like. Then just define this action of $G$ on $H$:   $g\cdot h=h$ when $h \notin Y.$ However $g \cdot y_i=y_j$ where the permutation $g$ sends the integer $i$ to the integer $j.$ In other words each integer $i$ is replaced by the element $y_i \in H.$

This construction works any time you have a group $G$ acting faithfully on a set $X$ and a subgroup $H$ of cardinality greater than or equal to that of $X.$ 

Let me get more specific. I'll stick to cyclic groups so let $\sigma\in G$ have infinite order and let $H=<\sigma>$ be the countable cyclic subgroup it generates. I'll discuss some unusual choices of $\sigma$ at the end, but the construction is indifferent.

For the enumeration of a countable subset of $H$ I will take $y_1=\sigma^3,y_2=\sigma^5,y_3=\sigma^8,y_4=\sigma^{13},\cdots$ i.e. the powers of $\sigma$ whose exponents are Fibonacci numbers greater than $2,$ in order of increasing size. (Other, weirder, constructive orders could also be given)

There is no reason not to take $Y$ to be all of $H,$ but also no reason to do so.


There are obvious nice choices for $\sigma$ but it works just as well to take

$\sigma=(\cdots 29\ 17\ 13\ 5\ 3\ 7\ 11\ 19 \cdots)$ which moves only the odd primes or

$\sigma=(10\ 11)(100\ 101\ 102)(1000\ 1001 \ 1002\ 1003)\cdots$ which has cycles of each order and moves only numbers of the form $10^i+j$ with $1 \le i $ and $0 \leq j \leq i.$ 

Finally, $H$ could really be any infinite subgroup. The only use made of the fact that it was cyclic was that I could easily list a sequence of distinct elements.