If $ι:U→V$ is a Hilbert-Schmidt embedding and $(v_n)_{n∈ℕ}$ is an orthonormal basis of $V$, then $(ιι^*v_n)_{n∈ℕ}$ is an orthonormal basis of $ιU$

Question

Let

• $(U,\langle\;\cdot\;,\;\cdot\;\rangle_U)$ and $(V,\langle\;\cdot\;,\;\cdot\;\rangle_V)$ be separable $\mathbb R$-Hilbert spaces
• $\iota:U\to V$ be a Hilbert-Schmidt embedding
• $T:=\iota\iota^\ast$
• $V_0:=T^{1/2}V$ be equipped with $$\langle u,v\rangle_{V_0}:=\langle T^{-1/2}u,T^{-1/2}v\rangle_V\;\;\;\text{for }u,v\in V_0\;,$$ where $$T^{-1}:=\left(\left.T\right|_{\ker(T)^\perp}\right)^{-1}$$ denotes the pseudo inverse of $T$

We can show that

1. $V_0=\iota U$ (in the sense of set equality)
2. $\iota$ is an isometry between $(U,\langle\;\cdot\;,\;\cdot\;\rangle_U)$ and $(V_0,\langle\;\cdot\;,\;\cdot\;\rangle_{V_0})$
3. $T$ is a bounded, linear, nonnegative and symmetric operator on $V$ with finite trace

Let $(v_n)_{n\in\mathbb N}$ be a suitable$^1$ orthonormal basis of $(V,\langle\;\cdot\;,\;\cdot\;\rangle_V)$. Can we show that $(T^{1/2}v_n)_{n\in\mathbb N}$ is an orthonormal basis of $(V_0,\langle\;\cdot\;,\;\cdot\;\rangle_{V_0})$?

If that's not possible, can we show it, if we assume that $U\subseteq V$ ($U$ and $V$ being still equipped with different inner products) and $\iota$ is the inclusion? Note that in that case $(V_0,\langle\;\cdot\;,\;\cdot\;\rangle_{V_0})=(U,\langle\;\cdot\;,\;\cdot\;\rangle_U)$.

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$^1$ It's not important for me whether or not we can show the statement for every or only for a special choice of $(v_n)_{n\in\mathbb N}$. A necessary special choice might be the choice for which we've got $$Tv_n=\mu_nv_n\;\;\;\text{for all }n\in\mathbb N\tag 1$$ for some $(\mu_n)_{n\in\mathbb N}\subseteq[0,\infty)$. The existence of such a $(v_n)_{n\in\mathbb N}$ is guaranteed by the Hilbert-Schmidt theorem.