Let $C_{\lambda,\mu}$ be the coefficients defined as $$ s_\lambda\left(\frac{x_1}{1-x_1},...,\frac{x_N}{1-x_N}\right)=\sum_{\mu\supset \lambda}C_{\lambda\mu}s_\mu(x_1,...,x_N),$$ where $s$ are the Schur polynomials.

These coefficients have an explicit form in terms of a determinant of binomial coefficients, $$ C_{\lambda\mu}=\det\left[\binom{\mu_j-j}{\lambda_i-i}\right].$$

Given a partition $\lambda$, the set $\{j-i,1\le j\le \lambda_i,1\le i\le \ell(\lambda)\}$ is its content set. Let $t_\lambda$ be the product of all non-zero contents of $\lambda$.

I have met the following identity which I believe to be true: $$ \sum_{\mu\supset\lambda}C_{\lambda\mu}\frac{t_\mu^2}{s_\mu(1^N)}=\frac{t_\lambda^2}{s_{\lambda'}(1^N)},$$ where $\lambda'$ is the partition conjugated to $\lambda$.

Notice the infinite amount of cancellation: Since $s_\mu(1^N)=0$ for $N<\ell(\mu)$, the left hand side might in principle be singular for all integer $N$, but the right hand side is actually only singular for $N<\ell(\lambda')$.

Has anyone seen this identity before?

Edit. Richard Stanley has mentioned that $C_{\lambda\mu}=\frac{|\lambda|!}{|\mu|!}f^{\mu/\lambda}\frac{T_{\mu/\lambda}}{t_\lambda}$, where $T_{\mu/\lambda}$ is the product of all contents of the skew partition $\mu/\lambda$. Then the identity can also be written as $$\sum_\mu\frac{f^{\mu/\lambda}}{|\mu|!}\frac{T_{\mu/\lambda}^3}{s_\mu(1^N)}=\frac{1}{|\lambda|!s_{\lambda'}(1^N)}$$ (here $f^{\mu/\lambda}$ is the number of standard Young tableaux of that skew shape).

Edit2. Let me give an example that will also help clarify the meaning of the equality. I have in mind a large $N$ expansion. Take the case $\lambda=(1,1)$. If I add the terms corresponding to $\mu\in\{(1,1),(2,1),(1,1,1)\}$ I get $$ \frac{N^2-3N-8}{N(N-2)(N^2-1)}$$ for the left hand side, which is $$ \frac{1}{N^2}- \frac{1}{N^3}+O\left(\frac{1}{N^4}\right)$$ for large $N$. If I include further values of $\mu$, more terms in the large $N$ expansion of the left hand side agree with $$\frac{1}{N^2}-\frac{1}{N^3}+\frac{1}{N^4}-\frac{1}{N^5}+\cdots,$$ which is the large $N$ series of $\frac{1}{N(N+1)}$, the right hand side.