The 1-D heat equation $u_t=\gamma u_{xx}$ admits the transformation
$$\frac{\partial}{\partial x}, \mbox{ translation along x}$$
The solution $u$ is invariant. This means it exhibits translational invariance. Then, if we transform it along $x$, we get
$$u(x,t) \rightarrow u(x-a,t)$$
Both are solutions, because symmetries "transform solutions to solutions". What does this mean, though. Does it mean
$$u(x,t)=u(x-a,t)?$$
because an invariant doesn't change under transformation? The definition of invariant is $u(x') \equiv u(x)$, or equivalently, if we use the infinitesimal generator of the transformation $\hat{U}$: $\hat{U}u=0$. Then it must be true that
$$u(x,t)=u(x-a,t)$$
But this would mean, that I can select $a$ arbitrarily and then $u$ is exactly the same anywhere along $x$. Then $u=\mbox{const}$, correct? This cannot be true. Where is the mistake? What is the meaning and what is the usefulness of this invariance.
It doesn't mean $u(x,t) = u(x-a,t)$, since then the solutions would be constants. Rather, it means that if you define another function $\psi(x,t) = u(x-a,t)$, then $\psi$ is also a solution. You note this above. THat's what it means.