I have already understand the basic proof of Catalan numbers through monotonic lattice paths. However, I am know figuring out the specific case: if the two paths are symmetric about a diagonal line (from lower-left to upper-right), they should be counted only once.
For example, the two matrices (the boundary between 0 and 1 in upper triangular matrix is the monotonic path), \begin{bmatrix}1&0&0\\&1&1\\&&1\end{bmatrix}\begin{bmatrix}1&1&0\\&1&0\\&&1\end{bmatrix} They are isomorphic by my definition, and should be counted only once. Thus, how to get the number of equivalent classes in this case? Thanks!
Let $C_n$ be the $n^\text{th}$ Catalan number, which is the total number of paths without regard to the reflection condition.
Let $R_n$ be the number of Catalan paths which are equal to their own reflection through the skew-diagonal.
Using Burnside's lemma, the number of non-equivalent up to reflection paths is equal to $$ \frac12(C_n+R_n). $$ There is a well-known formula for $C_n$, so you just need to find a formula for $R_n$. Note that a path counted by $R_n$ is completely determined by its first $n$ steps, because the remaining $n$ steps are the reverse-complement of the first $n$ steps. These first $n$ steps must be a sequence of $n$ symbols, each of which is either $\to$, for a right step, or $\downarrow$, for a downward step. This sequence is valid if and only if, for each initial segment, there are at least as many $\to$'s as $\downarrow$'s. Using this answer by Marc van Leeuwen, or my alternate solution, we conclude that $$ R_n=\binom{n}{\lfloor n/2\rfloor}. $$ Therefore, the number of paths up to reflection symmetry is $$ \frac12\left[\frac1{n+1}\binom{2n}n+\binom{n}{\lfloor n/2\rfloor}\right]. $$