Posterior of one observation transform into posterior of several observations Suppose $\mu$ has prior distribution $\mathcal{N}(M, A)$ and $x |\mu \sim \mathcal{N}(\mu, 1)$
After one observation, the posterior is $$\mu|x \sim \mathcal{N}(M + B(x-M), B), \tag{1}$$ where $B \overset{\text{(def)}}{=} \dfrac{A}{A+1}$.
After $n$ independent observations $\mathbf{x} \overset{\text{(def)}}{=}x_1, \dotsc, x_n | \mu\overset{\perp \!\!\perp}{\sim} \mathcal{N}(\mu, 1)$, the posterior is $$\mu|\mathbf{x} \sim \mathcal{N}\left(M + B_n(\bar{x} - M), \frac{1}{n}B_n \right), \tag{2}$$ where $B_n \overset{\text{(def)}}{=} \dfrac{nA}{nA + 1}$.
Is there the general method to calculate the posterior after several observations, given the prior and the posterior after one observation? Is there a path that can take me from <(1) , prior> to (2) in different situations?
It should be just manipulating Bayes' rule...someone help me out here. Thanks

Exponential families are notable precisely because there is a general method like the one you’re (I’m) searching for. Unfortunately this is somewhat obscured in the Wikipedia article.
Firstly in exponential families we are dealing with families of measures that are by definition absolutely continuous with respect to one another -- they are exponential tiltings of a given measure. So we are free to talk in terms of densities${}^1$ with respect to some fixed measure (maybe Lebesgue measure, for instance).
Suppose the prior for $\mu$ has density $g(\mu)$, and $x |\mu$ has density $f_{\mu}(x)$. Then the posterior on $x_1, \dotsc, x_n$ independent observations is $\mu|\mathbf{x}$ has density $\propto g(\mu) \prod_{i=1}^nf_\mu(x_i)$
Exponential families can simplify this. If we are in an exponential family, we can find the (unnormalized) posterior after $n$ observations with a simple adjustment, using no other information but the average $\bar{y} = \frac{1}{n}(y(x_1) + \dotsb + y(x_n))$ of the sufficient statistic.
In more detail, suppose the prior for $\mu$ has density $g(\mu)$, and $x |\mu$ has density $e^{\alpha y - \psi (\alpha)}f_{\mu_0}(x)$ for $y = t(x)$ and $\alpha = \alpha(\mu)$ the natural parameter. Then the posterior $\mu | \mathbf{x}$ has density $\propto g(\mu)e^{\alpha \left( \sum y_i \right) - n \cdot \psi(\alpha)} \prod f_{\mu_0}(x_i)$, and we can remove the $\prod f_{\mu_0}(x_i)$ since we are satisfying ourself with an un-normalized density, and $\prod f_{\mu_0}(x_i)$ does not depend on $\mu$. (The quantity $\mu_0$ is some fixed constant.)
$$\mu | \mathbf{x} \propto g(\mu)e^{\alpha \left( \sum y_i \right) - n \cdot \psi(\alpha)} \,\,\text{ AKA }\,\,g(\mu)e^{n \cdot (\alpha \bar{y} - \psi(\alpha))}$$
See here for another form.
Relevant is this from page 69 of Efron/Hastie's Computer Age Statistical Inference

In 1935–36, a trio of authors, working independently in different
countries, Pitman, Darmois, and Koopmans, showed that exponential
families are the only ones that enjoy ﬁxed-dimensional sufﬁcient
statistics under repeated independent sampling.


1: See I-projection, where an exponentially-tilted version of a measure $\mu$ is the orthogonal projection of $\mu$ in a certain inner product space, onto $\{\nu \mid \nu \ll \mu \text{ and }  C\}$, where $C$ is some constraint.
