Let $K$ be a compact Lie group and let $K$ act on $L^2(K)$ via the left regular representation. That is, for $\phi\in L^2(K)$ and $k, h\in K$,
$$(k\cdot\phi)(h):=\phi(k^{-1}h).$$
One knows (for example by the paper "Cyclic Vectors for Representations of Locally Compact Groups" by Greenleaf and Moskowitz 1971) that there exists a cyclic vector in $v\in L^2(K)$ - that is, a $v$ such that the linear span of the set $K\cdot v$ is dense in $L^2(K)$.
Let $P:L^2(K)\rightarrow\mathbb{C}v$ be the orthogonal projection onto the subspace spanned by $v$. Define a bounded linear map
$$T:L^2(K)\rightarrow L^2(K,\mathbb{C}v)$$ $$\phi\mapsto T(\phi),$$
where $T(\phi)$ is the function that maps $k\in K$ to $P(k^{-1}\cdot\phi).$
One can verify that, with respect to the left-regular representations on both sides, $T$ is a $K$-equivariant map. $T$ is also an isometry. I'm looking for clarification on the following points:
Is $T$ injective? I believe so, but I'd like some confirmation. The idea is that the map $T$ finds the value of $\phi$ at each left-translate of $v$, and since $v$ is cyclic, such translates are dense in $L^2(K)$. Thus if $T(\phi)=0$, then $\phi$ should be $0$.
Is $T$ surjective? For this, it seems important to know something about the set $A:=\{kv:k\in K\}$, in particular whether or not $A$ is linearly independent.