A group action contains another action as a normal subgroup

A group action contains another action as a normal subgroup

I have a question on what it means for one group action to contain another
group action as a normal subgroup. I am guessing this means that the image
of the first group action contains the image of the second as a normal
subgroup, but I want to double check.
The context where I saw this phrase is as follows. Given a group $G$, the
right and left regular actions of $G$ on itself are defined by the maps
$\rho: (x,g) \mapsto xg$ and $\lambda:(x,g) \mapsto g^{-1}x$,
respectively. Then, the action of $G^* = G \times G$ on $G$, which is the
product of the left and right regular actions, is defined by
$\mu(x,(g,h))=g^{-1}xh$. The text I'm reading says
``This transitive action contains both the left and right regular actions
as normal subgroups.''
My question is on what it means for one action to contain another as a
normal subgroup. Since an action on $G$ corresponds to a homomorphism into
$Sym(G)$, the image of this homomorphism (i.e. the permutation
representation afforded by this action) could be what was meant. In fact,
in this case, I verified that the image of the right regular action $\rho$
(and also of $\lambda$) is a normal subgroup of the image of the product
action $\mu$ (all three of these images are subgroups of $Sym(G)$). So I
am guessing what is meant by one action being a normal subgroup of another
is that its image is a normal subgroup of the other image. Is that
correct?
We can't always reduce or think of an action in terms of just its image
since we lose some information if the action is not faithful. For example,
two actions need not be isomorphic even if their images are. In fact, the
definition of isomorphic actions is in terms of the action and not just in
terms of the images of the actions.