= Solution
Use $v=\sigma^2$ as the <variance> parameter, with prior density relative to $d\mu\,dv$. Let $\bar x=n^{-1}\sum x_i$ and $S=\sum(x_i-\bar x)^2$. Multiplying the prior by the <likelihood function> gives the unnormalized <Bayesian posterior>
$$
\boxed{\pi(\mu,v\mid x)\propto v^{-n/2-1}\exp\left[-\frac{S+n(\mu-\bar x)^2}{2v}\right],\qquad v>0.}
$$
The printed request for an “improper posterior” should not be read as a claim that this posterior is always improper. Although the prior is improper, integrating over $\mu$ gives a marginal kernel $v^{-(n+1)/2}e^{-S/(2v)}$, which is integrable exactly when $n>1$ and $S>0$. Under these usual nondegenerate-data conditions it is a proper posterior. If those conditions fail, the displayed kernel is improper and there is no posterior probability distribution for a sampler to target.
Completing the square and using the <inverse-gamma distribution> give the full conditionals
$$
\boxed{\mu\mid v,x\sim N(\bar x,v/n),\qquad v\mid\mu,x\sim\operatorname{IG}\left(\frac n2,\frac{S+n(\mu-\bar x)^2}{2}\right).}
$$
The <inverse-gamma distribution> convention is density proportional to $v^{-\alpha-1}e^{-\beta/v}$. Equivalently the precision satisfies
$$
\boxed{\tau=1/v\mid\mu,x\sim\operatorname{Gamma}\left(\frac n2,\frac{S+n(\mu-\bar x)^2}{2}\right),}
$$
using shape and rate, as in the question.
A <Gibbs sampler> starts from $v^{(0)}>0$ and alternates
$$
\mu^{(k+1)}\sim N(\bar x,v^{(k)}/n),\qquad
\tau^{(k+1)}\sim\operatorname{Gamma}\left(n/2,\tfrac12\sum_i(x_i-\mu^{(k+1)})^2\right),\qquad
v^{(k+1)}=1/\tau^{(k+1)}.
$$
This is the <normal mean-variance posterior with a log-uniform variance prior>. For a marginal check, $v\mid x\sim\operatorname{IG}((n-1)/2,S/2)$: integrating out $\mu$ reduces the shape by $1/2$. This marginal shape is different from the full-conditional shape $n/2$.
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