Suppose we are trying to estimate a real valued parameter under the linear exponential (LINEX) loss. Suppose we have some prior $g$ that has finite mean, variance and a moment generating function.
What is the Bayes estimator without being given data?
What about when data is normal with mean $\theta$ (unknown) and variance 1, under a normal prior $g$? Here the real valued parameter is the mean.
For an arbitrary density $f(\theta)$, the expected LINEX loss is
$$ E[L(\theta, a)] = \int [e^{c(a - \theta)} - c(a - \theta) - 1] f(\theta) \ d\theta. $$
Differentiating with respect to $a$ shows that the minimum occurs when
$$ \int [c e^{c(a - \theta)} - c] f(\theta) \ d\theta = 0. $$
Solving this gives
$$ a = \frac{-\log \int e^{-c\theta} f(\theta) \ d\theta}{c} = \frac{-\kappa(-c)}{c} $$
where $\kappa(\cdot)$ denotes the cumulant-generating function.
I'm going to answer a modified version of your follow-up question where we assume $\theta$ has a flat prior and $X \sim N(\theta, 1)$. Then, $[\theta \mid X = x] \sim N(x, 1)$. The cumulant-generating function of an $N(x, 1)$ distribution is $\kappa(t) = xt + t^2 / 2$, so this gives
$$ a(X) = X - c / 2. $$
