Maps from a (f.g.) $k$-algebra (w/action of finite group $G$) to $k$, that agree on the subring of $G$-invariants Suppose $A$ is a finitely generated $k$-algebra with action of a finite group $G$. Suppose we have two maps $\Phi: A\to k$ and $\Phi': A\to k$ such that $\Phi|_{A^G}=\Phi'|_{A^G}$, i.e. they agree on the subring of $G$-invariant elements. I want to prove the following (I'm not positive it is true but pretty sure it should be)

There exists $g\in G$ such that for any $a\in A$ we have $\Phi'(a) = \Phi(g\cdot a)$ ($g$ is the same for all $a$).

I managed to do half the job: For any $a\in A$ one can construct the polynomial
$$
p_a(X) = \prod_{g\in G}(X-g\cdot a)
$$
The coefficients of this polynomial are all in $A^G$ and for all $h\in G$, we have $p_a(h\cdot a) = 0$. This means $A$ is integral over $A^G$. Inspired by the above polynomial construct
$$
q^{\Phi}_a(X) = \prod_{g\in G}(X-\Phi(g\cdot a)), \quad
q^{\Phi'}_a(X) = \prod_{g\in G}(X-\Phi'(g\cdot a))
$$
Since the coefficients of both polynomials are in the image of $A^G$ under $\Phi$ and $\Phi'$, and since $\Phi|_{A^G}=\Phi|_{A^G}$, these polynomials are the same. This means for any $a\in A$, there exists $g_a\in G$ such that $\Phi'(a) = \Phi(g_a\cdot a)$. This is because $\Phi'(a)$ is a root of $q_a^{\Phi}$.
Moreover since $A$ is integral over $A^G$ (both finitely generated), then $A$ is an $A^G$-module of finite type. Let $\{m_i\}_{i\leq N}$ be the generators of this $A^G$-module. Denote by $g_i\equiv g_{m_i}$ in the above construction. If one can prove that, there exists $g\in G$, such that for all $i\leq N$ we have $\Phi'(m_i)=\Phi(g_i\cdot m_i)=\Phi(g\cdot m_i)$, we are done.
I've spent hours trying to show this and came up empty. Any hints?
Edit: $k$ is a field (even algebraically closed) and $A$ is a commutative $k$-algebra.
 A: I figured it out. There is a one-to-one correspondence between a $k$-algebra homomorphism like $\Phi: A\to k$ and its kernel $m_\Phi=\mathrm{ker}(\Phi)$ which is a maximal ideal. If two maps like this have the same kernel then they are equal.
Now if $a\in \ker(\Phi')=m_{\Phi'}$, then there exists $g_a$ (as I proved before) such that $g_a\cdot a\in \ker(\Phi)=m_\Phi$. Therefore $a\in g^{-1}\cdot m_\Phi$. But for any $g\in G$ the ideal $g^{-1}\cdot m_\Phi$ is maximal since it is the kernel of $\Phi(g\cdot \bullet):A\to k$. Therefore $m_{\Phi'}\subset \bigcup_{g\in G} g^{-1}\cdot m_\Phi$.
Now we have the theorem that if $\mathfrak{a}$ is an ideal such that $\mathfrak{a}\subset\bigcup_{i= 1}^n \mathfrak{p}_i$ such that $\mathfrak{p}_i$ are prime then $\mathfrak{a}\subset \mathfrak{p}_j$ for some $j$. In our case then this means there exists $h\in G$ such that $m_{\Phi'} = h^{-1}\cdot m_\Phi$, or $\ker(\Phi') = \ker(\Phi(h\cdot\bullet))$. Which means $\Phi'(a) = \Phi(h\cdot a)$ for all $a\in A$. QED.
