Here is the sort of example I think Jason Starr was raising. (I looked at Brian's webpage, but it wasn't obvious which paper to read.) Take $k$ to be a perfect field of characteristic $p$, with $p \neq 0$, $2$. Let $A = k[x,y]/(y^2-x^p)$. The normalization of $A$ is $\tilde{A} = k[t]$, with $y=t^p$ and $x=t^2$. 

Let $G$ be the group scheme $\mathbb{G}_a$, so it has underlying space $k[\epsilon]/\epsilon^p$ and multiplication given by the map $\epsilon \to \epsilon \otimes 1 + 1 \otimes \epsilon$. If, like me, you prefer to think in terms of functors of points, $G(R) = \{ \epsilon \in R : \epsilon^p =0 \}$ and the multiplication map $G(R) \times G(R) \to G(R)$ is $(\epsilon_1, \epsilon_2) \mapsto \epsilon_1 + \epsilon_2$.

Let $G(R)$ act on $A(R)$ by $(\epsilon, (x,y)) \mapsto (x+2 \epsilon, y)$. If you prefer maps of algebras, $x \mapsto 1 \otimes x + 2 \epsilon \otimes 1$, $y \mapsto 1 \otimes y$. I claim that this action does not lift to $\tilde{A}$. Suppose, to the contrary, that $t \mapsto 1 \otimes t_0 + \epsilon \otimes t_1 + \cdots + \epsilon^{p-1} \otimes t_{p-1}$, with the $t_i \in k[t]$. Writing out that the action must preserve the relation $t x^{(p-1)/2} = y$ gives
$$ \left( 1 \otimes t_0 + \epsilon \otimes t_1 + \cdots \right) (1 \otimes t^2 + 2 \epsilon \otimes 1)^{(p-1)/2} = 1 \otimes t^p $$

Equating the coefficients of $1$ and $\epsilon$ gives $t_0 t^{p-1} = t^p$ and $t_1 t^{p-1} + (p-1) t_0 t^{p-3}  = 0$. So $t_0=t$ and $t_1 = 1/t$. But $1/t$ isn't in $k[t]$. 

Morally, the action wants to be $(\epsilon, t) \mapsto t \sqrt{1+2\epsilon t^{-2}}=1+\epsilon t^{-1} - (1/2) \epsilon^2 t^{-3} + \cdots $. The trouble is that $\left( t \sqrt{1+2\epsilon t^{-2}} \right)^k$ is in $k[t, \epsilon]/\epsilon^p$ when $k$ is even or is $\geq p$, but not in general.

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This can never occur when $G$ is normal. Over a field of characteristic zero, all group schemes are regular, and regular implies normal, so there are no examples over a field of characteristic zero.

Recall the universal property of normalization: For any normal variety $Y$, the induced map $\mathrm{Hom}(Y, \tilde{X}) \to \mathrm{Hom}(Y,X)$ is bijective.

Proof: If $G$ is normal, then $G \times \tilde{X}$ is normal. We have a map $G \times \tilde{X} \to  G \times X \to X$, where the first map is $\mathrm{Id} \times \phi$ and the second map is the group action. By the universal property of the normalization, there is a map $G \times \tilde{X} \to \tilde{X}$ making the obvious diagram commute. We claim that this map gives an action of $G$ on $\tilde{X}$.

Consider the two maps $G \times G \times \tilde{X} \to \tilde{X}$. We must show they are equal. Again, by the universal property of normalization, it is enough to show that the two compositions $G \times G \times \tilde{X} \to X$ are equal. But these are the same as $G \times G \times \tilde{X} \to G \times G \times X$, follwed by the two maps $G \times G \times X \to X$, and these are equal because $G$ acts on $X$.