Let $A=(a_{ij})\in\operatorname{SL}_2(\mathbb{F}_p)$. Consider the ring map $A:\mathbb{F}_p[x,y]\to\mathbb{F}_p[x,y]$ defined by
$$A(x)=a_{11}x+a_{21}y$$ $$A(y)=a_{12}x+a_{22}y$$
and extended multiplicatively. Are there non-constant polynomials $f(x_1,x_2)\in\mathbb{F}_p[x,y]$ such that $f(x,y)=A(f(x,y))$ for all $A\in\operatorname{SL}_2(\mathbb{F}_p)$?
I tried to solve this problem by writing $f(x,y)=\sum\lambda_{ij}x^iy^j$, and discovering different restrictions on the $\lambda_{ij}$ for particularly nice choices of $A$. This quickly got complicated and so I'm hoping this has been studied before and there is a more elegant solution.
Edit: After working through this for some small primes, I believe that the following two polynomials are fixed by all $A$:
$$f(x,y)=x^py-xy^p$$ $$g(x,y)=\sum_{i=0}^px^{(p-i)(p-1)}y^{i(p-1)}$$
However, I can't prove that $g(x,y)$ is fixed for all $p$, and it's not clear to me that these two polynomials generate the entire subalgebra of $\mathbb{F}_p(x,y)$ invariant under the action of $\operatorname{SL}_2(\mathbb{F}_p)$. Is there an invariant $h(x,y)$ not in the subalgebra generated by $f$ and $g$?
These are the Dickson's polynomials. If we let $$h(x,y)=x^{p^2}y-y^{p^2}x$$ then $h$ is also invariant under $SL(2,p)$ action. It then follows that $$g(x,y)=\frac{h(x,y)}{f(x,y)}$$ is invariant.
For a general $n$, let
$$L_{n,s}=\left|\begin{array}{cccc} x_1&x_2&\cdots&x_n\\ x_1^{p}&x_2^{p}&\cdots&x_n^p\\ \cdots&\cdots&\cdots&\cdots\\ \widehat{x_1^{p^s}}&\widehat{x_2^{p^s}}&\cdots&\widehat{x_n^{p^s}}\\ \cdots&\cdots&\cdots&\cdots\\ x_1^{p^n}&x_2^{p^n}&\cdots&x_n^{p^n} \end{array}\right|$$
where the $p^s$ row is omitted. Then the Dickson's invariant for $SL(n,p)$ are $\dfrac{L_{n,s}}{L_{n,n}}$.
And you are correct, these two generate the whole invariant algebra. I don't know the details but they are likely in Dickson's works.