= Solution
Both fits use the same <Gamma distribution> mean model and inverse <link function>. Their coefficient estimates, fitted means and hence their weight matrices coincide: the common <dispersion parameter> only multiplies the likelihood score by a scalar and therefore does not change its zero. The displayed calls likewise show the same family and formula; fixing dispersion in a summary changes the uncertainty calculation, not these coefficient estimates.
For $V(\mu)=\mu^2$ and $g'(\mu)=-\mu^{-2}$, the <Fisher information> weights are
$$
W_{ii}=\{a_i\mu_i^2\mu_i^{-4}\}^{-1}=\mu_i^2/a_i.
$$
Thus the common matrix $(X^TWX)^{-1}$ is multiplied by dispersion $1$ in the exponential case and by $\widehat\phi=0.3103711$ in the fitted gamma case. Taking square roots of the diagonal <covariances> gives
$$
\boxed{SE_{\mathrm{mod2}}(\widehat\beta_j)=\sqrt{0.3103711}\,SE_{\mathrm{mod1}}(\widehat\beta_j)\approx0.5571\,SE_{\mathrm{mod1}}(\widehat\beta_j).}
$$
For instance $0.03607\times0.5571\approx0.02009$ for the intercept, and the same factor applies to all the coefficients. The <standard errors> are smaller because the fitted <dispersion parameter> is below one, not because a different mean function was fitted.
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