Ideals invariant under ring automorphisms

There are tons of examples. Put $u = x^2+xy+y^2$ and $v = xy(x+y)$. Then $u$ and $v$ are $SL_2(\mathbb{F}_2)$ invariant. If $f(s,t)$ and $g(s,t)$ are homogenous polynomials with respect to the grading $\deg s = 2$, $\deg t=3$, then $\langle f(u,v), g(u,v) \rangle$ is a $SL_2(\mathbb{F}_2)$-invariant generated in the manner you describe. I claim that, furthermore, if $\mathbb{F}_2[s,t]/\langle f, g \rangle$ is finite then so is $\mathbb{F}_2[x,y] / \langle f(u,v), g(u,v) \rangle$.

For this claim, we just need to show that $\mathbb{F}_2[x,y]$ is module finite over $\mathbb{F}_2[u,v]$. It is enough to show that the generators $x$ and $y$ are integral over $\mathbb{F}_2[u,v]$. Indeed, they are both roots of the monic polynomial $$z^3 + u z + v.$$ (I found $u$ and $v$ by multiplying out the polynomial $(z+x)(z+y)(z+x+y)$.)

One can see that $I=(f_1,\dots, f_s)$ is $G$-invariant iff $\alpha\cdot f_i \in I$ for each $\alpha \in G$ and $i$. From this one can prove easily that if $I,J$ are $G$-invariant, then so is $IJ$ and $I+J$. Thus, for example, $(x,y)^n$ are invariant for all $n\geq 0$. One also have that $I=(x^{2^k}, y^{2^k})$ is invariant as pointed out by YCor. So you can generate many other examples. A complete classification seems difficult though.

For the case of two-generated ideals, one could use the above remark and the fact that "Frobenius commutes with linear change of variables" to show:

1) If $I=(f,g)$ is invariant then $J= (f^{2^k}, g^{2^k})$ is invariant.

2) If $I=(f,g)$ is invariant and $\deg(f)> \deg(g)$ then $J=(f^{2^k}, g^l)$ is invariant with $l\leq 2^k$.

For instance the ideal $(x^6, x^2+xy+y^2)$ is invariant.