= Solution
Introduce the <Euler differential operator> $D=x\partial_x$. The fourth-order spatial expression is $D^4u$. The moving boundary has similarity coordinate $\xi=1$, so its two conditions give
$$
v(1)=0,\qquad t^\alpha v'(1)=1\quad(t>0).
$$
Hence $\alpha=0$ and $v'(1)=1$. With $D_\xi=\xi\partial_\xi$, substitution in the equation yields
$$
(D_\xi^4-\beta D_\xi)v=0,\qquad v(\xi)\to0\quad(\xi\to\infty).
$$
This is a <Cauchy-Euler equation>; putting $r=\log\xi$ turns it into the constant-coefficient equation $V^{(4)}-\beta V'=0$. Its <characteristic polynomial> is $n(n^3-\beta)$.
If $\beta>0$, write $a=\beta^{1/3}$. The roots are $0,a,-a/2\pm i\sqrt3a/2$. Decay excludes the constant and growing modes, leaving
$$
v(\xi)=\xi^{-a/2}\left[A\cos\left(\frac{\sqrt3a}2\log\xi\right)+B\sin\left(\frac{\sqrt3a}2\log\xi\right)\right].
$$
The boundary conditions force $A=0$ and $B=2/(\sqrt3a)$. If $\beta<0$, writing $\beta=-a^3$ leaves only one decaying mode, $C\xi^{-a}$; $v(1)=0$ then forces the zero function, which fails the derivative condition. If $\beta=0$, the solutions are polynomials of degree at most three in $\log\xi$, and none can both decay and satisfy $v'(1)=1$.
Consequently all real self-similar solutions are the one-parameter family
$$
\boxed{\alpha=0,\quad\beta>0,\quad s(t)=t^\beta,\quad u(x,t)=\frac{2}{\sqrt3\,\beta^{1/3}}\left(\frac{x}{t^\beta}\right)^{-\beta^{1/3}/2}\sin\left[\frac{\sqrt3\,\beta^{1/3}}2\log\left(\frac{x}{t^\beta}\right)\right]}.
$$
The <decaying profile for a fourth-order dilation equation> oscillates in logarithmic distance. No positivity condition on $u$ was imposed, so these sign changes are permitted.
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