Denote by $a$ a group action of $G$ on a finite set $X$, and consider the representation $(\rho_a,\mathbb{F}X)$ defined by $(\rho_a)_g(\sum_{x\in X}a_xx)=\sum_{x\in X}a_x(g.x)$. Prove that the dimension of $\mathbb{F}X^G=\{v\in\mathbb{F}X:\forall g\in G,(\rho_a)_g(v)=v\}$ is equal to the number of orbits of $a$.
To be clear, $\mathbb{F}X$ is our notation to the $\operatorname{Span}_{\mathbb{F}}\{x\in X\}$ - formal sums of elements in $X$. First I tried to understand when an element is in $\mathbb{F}X^G$ - $\sum_{x\in X}a_xx\in \mathbb{F}X^G\iff \forall g\in G, (\rho_a)_g(\sum_{x\in X}a_xx)=\sum_{x\in X}a_x(g.x)=\sum_{x\in X}a_xx$ and I think that $\mathbb{F}X^G=\operatorname{Span}\{x\in X:g.x=x\}$, so I tried maybe using Burnside lemma - but I think this is wrong since $\{x\in X:g.x=x\}$ maybe empty (for example a free action) but every action has atleast $1$ orbit.
Any hint would be helpful
Note that for each $g\in G$ we have:
$\rho(g)(\sum\limits_{x\in X}a_xx)=\sum\limits_{x\in X} a_x(g.x)=\sum\limits_{x\in X}a_{g^{-1}x}x$
Since the sums are formal, this element is equal to $\sum\limits_{x\in X} a_xx$ if and only if we have $a_x=a_{g^{-1}x}$ for all $x\in X$.
So the sum will belong to $\mathbb{F}X^G$ if and only if the following holds: if two elements of $X$ belong to the same orbit then their coefficients must be equal. And now there is a very natural way to construct a basis of $\mathbb{F}X^G$, which will prove your claim.