Banach fixed point theorem says that if $T$ is a mapping and there is a $q \in [0,1)$ where $|Tx - Ty| \leq q |x-y|$ for all $x,y$ then $T$ has a fixed point.
But if $T$ is a linear operator, then the $|Tx-Ty| \leq q|x-y|$ condition implies
$$|T(x-y)|\leq q|(x-y|,$$
which in turn implies $|Tz| \leq q|z|$ for any $z$ (since any vector $z$ can be represented as a difference of some pair of vectors $x$ and $y$).
But this seems to contradict $T$ having a fixed point, since if $Tz = z$, then $|Tz| = |z|\gt q|z|$ (unless $z=0$, but that would mean that only $0$ can be a fixed point).
What am I missing here?
Every linear mapping of the vector space to the same vector space has a fixed point (the origin). You don't need Banach's fixed point theorem for existence.
Also, Banach's theorem gives uniqueness if $q\in [0,1)$. However, the identity map has operator norm $1$, i.e $q\geq \|T\|=1$, which is clearly not in $[0,1)$, so the theorem cannot be applied, so you cannot deduce that there is a unique fixed point (also clearly for the identity map every point is fixed).