My knowledge on this topic is solely based on physics books, introductory texts on differential geometry and this one paper $[1]$.
One of the formulas of the paper shows what is called the Ricci soliton equation:
$R_{\mu\nu} -\frac{1}{2}\mathcal{L}_{X}g_{\mu\nu} - cg_{\mu\nu} = 0 \tag{1}$
In physics, a similar tensor field can be constructed: the Einstein tensor, given by:
$R_{\mu\nu} -\frac{1}{2}Rg_{\mu\nu} + \Lambda g_{\mu\nu} = 0 \tag{2}$
I would like to know: is it possible to have a situation where $\mathcal{L}_{X}g_{\mu\nu} =Rg_{\mu\nu}$, and $c=-\Lambda$ ?
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A vector field $X$ is said to be a conformal Killing vector field if ${\cal L}_Xg=\varphi g$ some function $\varphi$. In special case $\varphi=0$ it is called a Killing vector field.
Consider an Einstein Riemannian manifold $(M,g)$ of zero scalar curvature $R=0$ thus it is Ricci flat i.e. $Ric=R g=0$. If $(M,g)$ admit a Killing vector field $X$ (i.e. $\mathcal{L}_Xg=0$) then $(M, g, X)$ satisfies in your conditions. That is $0=\mathcal{L}_{X}g_{\mu\nu} =Rg_{\mu\nu}=0$ and $c=-\Lambda$ for zero Cosmological constant $\Lambda=0$.
For example $\Bbb R^n$ with Euclidean metric satisfies in your requirements. another example is torus $\Bbb S^1\times \Bbb S^1$ with flat metric because it admit Killing vector field.
All of these are correct if cosmological constant were zero, that I don't know what it is intuitively.