= Solution
The transformations
$$
g^s:(X,T,u)\longmapsto(e^sX,e^{-s}T,u)
$$
form a <one-parameter subgroup>[one-parameter group] with infinitesimal generator
$$
V=X\partial_X-T\partial_T.
$$
Under this scaling, $\partial_X$ has weight $-1$ and $\partial_T$ has weight $1$, so $u_{XT}$ has weight zero, as does $\sin u$. Thus $u_{XT}=\sin u$ is invariant and $g^s$ is a <Scaling symmetry of the sine-Gordon equation>.
The product
$$
z=XT
$$
is invariant under the group. A <group-invariant solution> therefore has the form $u(X,T)=F(z)$. Since
$$
u_X=TF'(z),
\qquad
u_{XT}=F'(z)+zF''(z),
$$
the <Sine-Gordon equation> reduces to
$$
zF''+F'=\sin F.
$$
Now put $w=e^{iF}$. Then
$$
F'=-i\frac{w'}w,
\qquad
F''=-i\left(\frac{w''}w-\frac{(w')^2}{w^2}\right),
\qquad
\sin F=\frac{w-w^{-1}}{2i}.
$$
Substitution and multiplication by $iw$ give
$$
zw''-z\frac{(w')^2}{w}+w'
=\frac{w^2-1}{2}.
$$
Hence
$$
\boxed{
w''=\frac{(w')^2}{w}-\frac{w'}z+\frac{w^2-1}{2z}}.
$$
This is the <Painlevé III equation> with parameters $(\alpha,\beta,\gamma,\delta)=(1/2,-1/2,0,0)$.
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