Breeder's equation and equivalent expressions for narrow-sense heritability
I found my mistake, which is where I said "To get the terms to cancel, we want to write $E$ as $\alpha E$ where $\alpha E$ has the same distribution as $G$." What we actually want is to find a constant $\alpha$ such that $\mathbb E[\alpha E\mid X=a] = \mathbb E[G\mid X=a]$. In my question, I assumed that this was the same thing as finding $\alpha$ such that $\mathbb E[\alpha E] = \mathbb E[G]$; hence I was trying to find $\alpha$ to make the distributions of $\alpha E$ and $G$ the same (the means are both 0, so I just needed to get the variances to equal each other). But actually, conditioning on $X=a$ shifts the distributions so the $\alpha$ I found in my question doesn't work.

Since my previous strategy doesn't work, I ended up skipping the step of finding $\alpha$ entirely. I realized that the random vector $(G, X)=(G,G+E)$ is a bivariate normal, because we can write $$\begin{pmatrix}G \\ G+E\end{pmatrix} = \begin{pmatrix}\sigma_G &amp; 0\\ \sigma_G &amp; \sigma_E\end{pmatrix}\begin{pmatrix}Z_1 \\ Z_2\end{pmatrix}$$ where $Z_1,Z_2$ are standard normal random variables.

Now Wikipedia gives the conditional distribution for bivariate normals, which in our case is $G\mid X=a$. We only need the mean, which is $\mu_G + \frac{\sigma_G}{\sigma_X}\rho (a - \mu_X)$. Both $\mu_G$ and $\mu_X$ are zero, and $\rho$ is the correlation which is $$\rho = \rho_{G,X} = \frac{\mathrm{cov}(G,X)}{\sigma_G \sigma_X} = \frac{\mathrm{cov}(G,G) + \mathrm{cov}(G,E)}{\sigma_G \sigma_X} = \frac{\sigma_G^2}{\sigma_G \sigma_X} = \frac{\sigma_G}{\sigma_X}$$ Thus we end up with $$\mu_G + \frac{\sigma_G}{\sigma_X}\rho (a - \mu_X) = \frac{\sigma_G^2}{\sigma_X^2}a$$ which is exactly what we want.