I recall coming across the equation $Rg_{ij}=\frac{1}{n(n-1)}R_{ij}$ at some point, and attempted to prove it today but was unsuccessful.
What I try: I use the notation in do Carmo's Riemannian Geometry.
First, I have $$ R= \frac{1}{n(n-1)}\sum_{ij} g^{ij}R_{ij} $$ Therefore, $$ Rg_{ab}= [\frac{1}{n(n-1)}\sum_{ij} g^{ij}R_{ij}]g_{ab} dx^a\otimes dx^b $$ To prove that $Rg_{ij}=\frac{1}{n(n-1)}R_{ij}$, we need to demonstrate that $(\sum_{ij} g^{ij}R_{ij})g_{ab}=R_{ab}$. However, $\sum_{ij} g^{ij}R_{ij}$ does not involve matrix multiplication, so we cannot use the analogue of $g_{kj}g^{ik}R_{ik}=\delta^i_jR_{ik}=R_{jk}$. Therefore, I am unsure how to proceed with this proof.
PS(2023/5/10): Inspired by Albanese's answer, I calculate the case $n=2$, and it is right. The calculation:
Under normal coordinate, namely $g_{ij}=\delta_{ij}$, since $n=2$, we have $$ R=\frac{1}{2}\sum_{ij}g^{ij}R_{ij}=\frac{1}{2}(R_{11}+R_{22}) $$ where $R_{11}=(e_1,e_2,e_1,e_2)=(e_2,e_1,e_2,e_1)=R_{22}$ ($\{e_1,e_2\}$ is the tangent vector of normal coordinate). Therefore, we have $$ R=R_{11}=R_{22} $$ Besides, $R_{21}=R_{12}=(e_1,e_2,e_2,e_2)+(e_1,e_1,e_2,e_1)=0$. So, we have $$ Rg_{11}=R=R_{11}, ~~~Rg_{22}=R=R_{22},~~~ Rg_{12}=Rg_{21}=0=R_{21}=R_{12} $$ namely,$Rg_{ij}=\frac{1}{2}R_{ij}$.
The identity you are trying to establish is false (consider a metric which has non-zero Ricci curvature but has scalar curvature zero).
If $(M, g)$ is a Riemannian manifold with $R_{ij} = fg_{ij}$ for some function $f$ and $\dim M \geq 3$, then it follows from the contracted differential Bianchi identity that $f$ must be constant, see Proposition $7.18$ of Lee's Introduction to Riemannian Manifolds (second edition). Denoting the constant by $\lambda$, we have $R_{ij} = \lambda g_{ij}$. Such a metric is called an Einstein metric. Taking the trace of this equation (and using Do Carmo's convention), we obtain $n(n-1)R = \lambda n$ and hence $\lambda = (n-1)R$. Therefore, if $g$ is an Einstein metric, it satisfies $Rg_{ij} = \frac{1}{n-1}R_{ij}$.