Consider what is sometimes known as generalized Catalan sequence $$\mathcal{{\color{red}C}}_{a,b}:=\frac{2b+1}{a+b+1}\binom{2a}{a+b}.$$ Observe that $\mathcal{{\color{red}C}}_{n,0}$ reduces to the usual Catalan number $C_n=\frac1{n+1}\binom{2n}n$.
Now, introduce a notion of convolution between $\mathcal{{\color{red}C}}$ and $C$, with respect to a kernel $\mu$, as follows: $$\mathcal{{\color{red}C}} *_\mu C:=\sum_{k=m}^n\mu_{n,m}(k)\,\mathcal{{\color{red}C}}_{k,m}\, C_{n-k}.$$
Let's look at the result of choosing three different kernels: \begin{align*} \sum_{k=m}^n \frac{(2m+1)(n-k+1)}{2n+1}\,\mathcal{{\color{red}C}}_{k,m}\, C_{n-k}&=\mathcal{{\color{red}C}}_{n,m}, \\ \sum_{k=m}^n \frac{m(m+k+1)}{k(2m+1)}\,\mathcal{{\color{red}C}}_{k,m}\, C_{n-k}&=\mathcal{{\color{red}C}}_{n,m}, \\ \sum_{k=m}^n \frac{m+k+1}{2n+1}\,\mathcal{{\color{red}C}}_{k,m}\, C_{n-k}&=\mathcal{{\color{red}C}}_{n,m}. \end{align*} In a sense, $C_n=\mathcal{{\color{red}C}}_{n,0}$ is serving as an identity $\mathcal{{\color{red}C}}*_{\mu}C=\mathcal{{\color{red}C}}$ w.r.t. the three different kernels.
QUESTION. Are there other (non-trivial) kernels satisfying the same property? What is the complete family?
Remark. Once a kernel is found, proving the identities is doable using the Wilf-Zeilberger machinery.
