No. In three dimensions there are three position operators, $\hat{x}_1$, $\hat{x}_2$, and $\hat{x}_3$, or maybe $\hat{x}$, $\hat{y}$, and $\hat{z}$. Each of these is a linear operator in the first, correct sense. Each one maps states in the Hilbert space onto other maps in the Hilbert space and nothing more.
Now, the three distinct position operators are actually closely related to each other, so we often write $\hat{\vec{x}}$ as a shorthand for talking about all of them at once, but they are still three separate operators that map $H\rightarrow H$.
Also, when you have three related position operators the Hilbert space does carry additional structure compared to when you have just one position operator. Specifically, the Hilbert space becomes $L_2(\mathbb{R}^3,\mathbb{C})$, rather than just $L_2(\mathbb{R},\mathbb{C})$. This is a larger Hilbert space. (At least in a certain sense. It is probably not the case that $L_2(\mathbb{R}^3,\mathbb{C}) \simeq L_2(\mathbb{R},\mathbb{C})^{\otimes 3}$ but it's an okay start)
Ultimately what I think you are looking for is the special relationship between $\hat{x}$, $\hat{y}$, and $\hat{z}$ that justifies grouping them together. The answer is symmetry. The Hilbert space $H = L_2(\mathbb{R}^3,\mathbb{C})$ transforms under rotations of physical space, represented by the group $SO(3)$. These rotations transform states in $H$ into other states in $H$, and they also transform operators on $H$ into other operators on $H$. For example, a rotation can map the $\hat{x}$ operator to the $\hat{y}$ and $\hat{y}$ to $-\hat{x}$, corresponding to a 90 degree rotation of your coordinate system.
When we call $\vec{\hat{x}}$ a vector operator we are really saying that the three operators transform into each other under rotations in this way. But there are still three distinct operators that individually map from $H$ to $H$!