In section 12.4.1 of Probability Theory: The Logic of Science by E.T. Jaynes, he describes the creation of a probability distribution $p(x|v\sigma) = \phi(x, \nu, \sigma) dx$ that is invariant under translation of the location parameter $\nu$ by $b$ or rescaling of the scale parameter $\sigma$ by $a$.
That is, the following functional equation:
$$\phi(x,\nu,\sigma)dx = \phi(x^{'},\nu^{'},\sigma^{'})dx^{'}$$
Holds under the transformations:
\begin{align} \nu^{'} & = \nu + b \\ \sigma^{'} & = a\sigma \\ x^{'} - \nu^{'} & = a(x - \nu) \end{align}
For all $0 < a < \infty$ and $-\infty < b < \infty$.
This produces: $$ \phi(x, \nu, \sigma) = a\phi(a(x-\nu) + \nu + b, \nu + b, a\sigma) $$
Jaynes suggests differentiating this functional equation with respect to a and b, and presents the solution as:
$$\phi(x,\nu,\sigma) = \frac{1}{\sigma}h\left(\frac{x - \nu}{\sigma}\right)$$
For an arbitrary function $h(q)$. Verifying that this is a solution is easy, however I was unable to derive it directly.
Differentiating with respect to a and b, as Jaynes suggested, produces the following PDEs:
\begin{align} 0 &= \phi(x,\nu,\sigma) + (x-\nu)\frac{\partial\phi}{\partial x} + \sigma\frac{\partial\phi}{\partial\sigma} \\ 0 &= \frac{\partial\phi}{\partial x} + \frac{\partial\phi}{\partial \nu} \end{align}
At this point I was unsure how to proceed and attempted to apply the method of characteristics, to no avail. I suspect that this is not the correct approach as I do not see how it could introduce the arbitrary function $h(q)$ into the solution.
Do the change of variables $$u=x-\nu,v = x+\nu \Leftrightarrow x = (v+u)/2, \nu = (v-u)/2, $$ with $\sigma$ the same (that is, if $s=\sigma$, then $\partial u/\partial s = \partial \nu/\partial s = 0$, so we don't need to bother with it).
Then $$0= \frac{\partial \phi}{\partial x}+\frac{\partial \phi}{\partial \nu}=\frac{\partial \phi}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial \phi}{\partial v}\frac{\partial v}{\partial x}+\frac{\partial \phi}{\partial u}\frac{\partial u}{\partial \nu}+\frac{\partial \phi}{\partial v}\frac{\partial v}{\partial \nu}=2\frac{\partial \phi}{\partial v}, $$ whence $$\phi = f(u,\sigma) = f(x-\nu,\sigma).$$
Now the first equation reads $$0 = f + u\frac{\partial f}{\partial u}+\sigma\frac{\partial f}{\partial \sigma}. $$
Do $w=u/\sigma,s=\sigma \Leftrightarrow u = ws, \sigma=s$ and let $f = g(w,s)/s$. Then $$\frac{\partial f}{\partial u} = \frac{\partial }{\partial w}\left(\frac{1}{s}g\right)\frac{\partial w}{\partial u}+\frac{\partial }{\partial s}\left(\frac{1}{s}g\right)\frac{\partial s}{\partial u}=\frac{1}{s^2}\frac{\partial g}{\partial w}$$ so $$u\frac{\partial f}{\partial u} = \frac{w}{s}\frac{\partial g}{\partial w}.$$
Furthermore, $$\frac{\partial f}{\partial \sigma} = \frac{\partial }{\partial w}\left(\frac{1}{s}g\right)\frac{\partial w}{\partial \sigma}+\frac{\partial }{\partial s}\left(\frac{1}{s}g\right)\frac{\partial s}{\partial \sigma}=-\frac{w}{s^2}\frac{\partial g}{\partial w}-\frac{1}{s^2}g+\frac{1}{s}\frac{\partial g}{\partial s}$$
so
$$\sigma\frac{\partial f}{\partial \sigma} = -\frac{w}{s}\frac{\partial g}{\partial w}-\frac{1}{s}g+\frac{\partial g}{\partial s}=-\frac{w}{s}\frac{\partial g}{\partial w}-f+\frac{\partial g}{\partial s}.$$
Thus $$0 = f+\frac{w}{s}\frac{\partial g}{\partial w} -\frac{w}{s}\frac{\partial g}{\partial w}-f+\frac{\partial g}{\partial s}, $$ so $$g = h(w) = h\left(\frac{x-\nu}{\sigma} \right)$$ and $$f = \frac{1}{\sigma} h\left(\frac{x-\nu}{\sigma} \right),$$ as desired.