= Solution
This is <attenuation bias from classical measurement error>, also called regression dilution. The latent predictor is $Z$, but $X_1$ and $X_2$ are noisy proxies. The noisier proxy produces a slope closer to zero.
For a regression with an intercept, the population slope is
$$
\frac{\operatorname{Cov}(X,Y)}{\operatorname{Var}(X)}.
$$
Here all variables are centred and the noises are independent. Therefore
$$
\operatorname{Cov}(Z,Y)=\beta\sigma_z^2,
\qquad
\operatorname{Var}(Z)=\sigma_z^2,
$$
so regressing $Y$ on $Z$ has slope $\beta=2$. For
$$
X_j=Z+\eta_j,
$$
we have
$$
\operatorname{Cov}(X_j,Y)=\beta\sigma_z^2,
\qquad
\operatorname{Var}(X_j)=\sigma_z^2+\sigma_{x_j}^2.
$$
Thus
$$
\boxed{
\beta_{\rm slope}(Y\sim X_j)
=\beta\frac{\sigma_z^2}
{\sigma_z^2+\sigma_{x_j}^2}
}.
$$
With the stated values, the slopes are
$$
2\frac1{1+0.5^2}=1.6,
\qquad
2\frac1{1+1^2}=1,
$$
matching the simulation.
Under this independent additive measurement-error model, the magnitude of the $Y$-on-$X_1$ slope is generally smaller than the magnitude of the $Y$-on-$Z$ slope whenever $\sigma_{x_1}^2>0$. Increasing $\sigma_y$ does not change the population slope, because response noise contributes neither to $\operatorname{Cov}(X_1,Y)$ nor to $\operatorname{Var}(X_1)$. It increases residual variance and the sampling variability of the estimate; with one million observations, doubling it should leave the displayed slope close to $1.6$ while increasing its standard error.
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