= Solution
For $N=1$, write
$$
q=\exp\left(-2lX-\frac{T}{2l}\right).
$$
The given evolution of the <discrete scattering data> gives
$$
\lambda_1=il,
\qquad
c_1(T)=2l\exp\left(-\frac{T}{2l}\right).
$$
Solving the resulting two scalar linear equations for the components of $\psi_1$ and substituting in the reconstruction formula gives
$$
u_X=-\frac{8lq}{1+q^2}.
$$
Choosing the additive multiple of $2\pi$ so that $u\to0$ as $X\to+\infty$ and <integral>[integrating] in $X$ yields the <One-soliton solution of the sine-Gordon equation in light-cone coordinates>
$$
\boxed{u(X,T)=4\arctan\exp\left(-2lX-\frac{T}{2l}\right)}.
$$
It depends only on $X+T$ exactly when the two positive coefficients in the exponent agree:
$$
2l=\frac1{2l}.
$$
The unique positive solution is $l=1/2$, and then
$$
u(X,T)=F(X+T),
\qquad
\boxed{F(z)=4\arctan(e^{-z})}.
$$
The transformations satisfy $g^s g^r=g^{s+r}$, $g^0$ is the <identity map>, and $(g^s)^{-1}=g^{-s}$, so they form a <group>[one-parameter group]. If
$$
U(X,T)=u(e^{-s}X,e^sT),
$$
then the <chain rule> gives
$$
U_{XT}=e^{-s}e^s u_{XT}=\sin U.
$$
Thus $g^s$ is a <Lie point symmetry>. Applied to the one-soliton family, it replaces $l$ by $l'=e^{-s}l$. Taking $e^s=2l$ gives $l'=1/2$, so every member is transformed to the function $F(X+T)$ found above.
For the stated $N=2$ solution, set $x=X+T$ and $t=T-X$. When $l^2+m^2=1/4$, its two arguments reduce to
$$
2mX-\frac{2mT}{4(l^2+m^2)}=-2mt,
\qquad
\frac{2lT}{4(l^2+m^2)}+2lX=2lx.
$$
At fixed $x$, only $\sin(2mt)$ varies, so the <Sine-Gordon breather> has fundamental period
$$
\boxed{\mathcal T=\frac{\pi}{m}}.
$$
Finally put $r^2=l^2+m^2$. Under the same symmetry, the parameters become
$$
l'=e^{-s}l,
\qquad
m'=e^{-s}m,
$$
and hence $(l')^2+(m')^2=e^{-2s}r^2$. Choosing $e^s=2r$ makes this sum $1/4$, proving that every solution in the family is symmetry-equivalent to the normalized breather.
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