Let $(E,\mathcal E)$ be a measurable space and $\kappa$ be a Markov kernel with density $p$ with respect to a measure $\lambda$ on $(E,\mathcal E)$; i.e. $$\kappa(x,B):=\int_Bp(x,y)\;\;\;\text{for }(x,B)\in E\times\mathcal E).$$ Note that $$(\kappa^\ast g)(y):=\int\lambda({\rm d}x)p(x,y)g(x)\;\;\;\text{for }y\in E$$ is well-defined outside a $\lambda$-null set for all $g\in\mathcal L^1(\lambda)$. $\kappa^\ast$ is a bounded linear operator on $\mathcal L^1(\lambda)$ and $$\int(\kappa f)g\:{\rm d}\lambda=\int f\kappa^\ast g\:{\rm d}\lambda$$ for all bounded $\mathcal E$-measurable $f:E\to\mathbb R$ and $g\in\mathcal L^1(\lambda)$.
Now let $\nu$ be a measure on $(E,\mathcal E)$ with density $q$ with respect to $\lambda$. Assume $\nu$ is $\kappa$-invariant; i.e. $\nu\kappa=\nu$; i.e. $$\int\nu({\rm d}x)\kappa(x,B)=\nu(B)\;\;\;\text{for all }B\in\mathcal E.$$
How can we show that $$\kappa^\ast(fq)=fq$$ for all bounded $\mathcal E$-measurable $f:E\to\mathbb R$?
My idea was to consider a "test function" $g:E\to\mathbb R$, bounded and $\mathcal E$-measurable, and observe $$\int(\kappa^\ast(fq)-fq)g\:{\rm d}\lambda=\int(\kappa f-f)gq\:{\rm d}\lambda=\int(\kappa f-f)g\:{\rm d}\nu\tag1.$$ However, how do I conclude from here? We would need to show that $(1)$ is $0$. This is only clear for me, when $g=1$.
This is false. For example, take $\lambda$ equal to counting measure on $E = \mathbb{Z}/N\mathbb{Z}$, and $p(x,y) = \mathbb{1}_{y=x+1}$ for $x,y \in E$. Then $\nu$ is the uniform measure on $E$, so $q$ is constant equal to $1/N$. For every $g : E \to \mathbb{R}$ and $y \in E$, we have $\kappa^*g(y)=g(y-1)$, so $\kappa^*$ is not the identity map.