Assume that we are in characteristic $0$. Let $G$ be an algebraic group acting on a quasi-projective variety $Y$. Let $G_0$ be the connected component of $G$ that contains the identity element. Let $P \in Y$ be an element and $\mathcal{O}$ the orbit of $P$ under the action of $G_0$, so $\mathcal{O}_0 = G_O \cdot P$. Let ${\mathcal{O}} = G \cdot P$ be the orbit of $P$ under the action of $G$. Assume that both $\mathcal{O}_0$ and ${\mathcal{O}}$ are quasi-projective sets. Is there anything we can say about the dimension of ${\mathcal{O}}$ when we know the dimension of $\mathcal{O}_0$?
I'm thinking of something like the following: We can write $G$ as a finite union of cosets: $$ G = \bigcup_{i=1}^n g_i G_0.$$ So ${\mathcal{O}} = \bigcup g_i \mathcal{O}_0$. If for some reason we had a disjoint union here, I think I would be able to show that $\mathcal{O}$ and $\mathcal{O}_0$ have the same dimension. But I think for a disjoint union we would have to assume that the stabilizer of $P$ is in $G_0$.
You have the right idea. Since $G = \bigcup_{i=1}^{n} g_i G_0$, we have $$ \mathcal{O} = \bigcup_{i=1}^{n} g_i \mathcal{O}_0. $$ The action of $g_i$ carries $\mathcal{O}_0$ isomorphically onto $g_i\mathcal{O}_0$, so these subsets of $X$ have the same dimension. The rest of the claim follows from topology. Indeed, suppose that $Z_1, \dots, Z_n \subset X$ are subspaces of dimension $n$ of a space $X$. Then $Z = \bigcup_{i=1}^{n} Z_i$ has dimension $n$ as well. Here's the proof. If $W_1 \subset W_2 \subset \dots \subset W_k$ is any ascending chain of closed, irreducible subsets of $Z$, then $$ W_k = \bigcup_{i=1}^{n} W_k \cap Z_i. $$ Since $W_k$ is irreducible and this union is finite, we have $W_k \cap Z_i = W_k$ for some $i$, i.e. $W_k \subset Z_i$. Therefore the entire chain is contained in $Z_i$, so $k \leq \text{dim}(Z_i) = n$, hence $\text{dim}(X) \leq n$. The converse inequality is immediate, since an ascending chain of closed irreducible subsets of any $Z_i$ defines such a chain in $X$.