Let $\mathcal{B(H)}$ be the space of all bounded linear operators on the Hilbert space $\mathcal{H}$. Let $g \rightarrow \mathcal{U}_g$, where $\mathcal{U}_g (\cdot):= U(g) (\cdot) U(g)^\dagger $, be the unitary representation of a group $G$ on $\mathcal{B(H)}$.
Let $\{ X_j^{(\lambda,\alpha)}\} \in \mathcal{B(H)}$ be such that \begin{align} \forall g \in G: \ \mathcal{U}_g[X_j^{(\lambda,\alpha)}] = \sum_{j'} U^\lambda_{j'j} (g) X_{j'}^{(\lambda,\alpha)}, \end{align} where $\lambda$ labels an irrep of $G$, $j$ labels the basis vector of the irrep $\lambda$, and $\alpha$ labels any multiplicity degrees of freedom, and \begin{align} U^\lambda_{j'j} (g) := \langle \lambda,j'| U^\lambda(g) |\lambda,j\rangle , \end{align} are the matrix elements of $U^\lambda(g)$, a unitary irreducible representation of $G$ on $\mathcal{H}$. The $\{X_j^{\lambda,\alpha}\}$ are then called an \emph{irreducible tensor operator basis} of $G$ on $\mathcal{B(H)}$.
We choose this basis to be normalized such that \begin{align} \text{Tr}[ {X_j^{(\lambda,\alpha)\dagger}} X_{j'}^{(\lambda',\alpha')} ] = \delta_{\lambda,\lambda'} \delta_{j,j'} \delta_{\alpha,\alpha'}. \end{align}
Is it possible to choose an irreducible tensor operator basis such that the singular values of of $X^{(\lambda,\alpha)}_j$ are all $\frac{1}{\sqrt{d_\lambda}}$, where $d_\lambda$ is the dimension of the irrep $\lambda$?
Here is a tentative answer of no. Please comment if I am misunderstanding the question. My claim is that $G = SU(2)$ has a counterexample. I will first motivate the counterexample, and then give a proof that it is the only allowed tensor operator basis.
Motivation:
For $\lambda$ the adjoint rep, also called the spin-$1$ representation, one has families of tensor operators for every rep of $SU(2)$: $X_{j=-1}^{(\lambda=adj)} = S^-, X_{j=0}^{(\lambda=adj)} = \sqrt{2} S^z, X_{j=1}^{(\lambda=adj)} = -S^+$. These are normalized for the fundamental representation on $\mathcal{H} = \mathbb{C}^2$, and can be normalized simply for other representations.
Explicitly for $\mathcal{H} = \mathbb{C}^2$, one would have $S^- = \begin{bmatrix}0 & 0\\ 1 & 0\end{bmatrix}$, and $S^z = \begin{bmatrix}1/\sqrt{2} & 0\\ 0 & -1/\sqrt{2}\end{bmatrix}$, and $-S^+ = \begin{bmatrix}0 & -1\\ 0 & 0\end{bmatrix}$. These satisfy your orthonormalization condition for $\lambda = \lambda' = \text{adjoint}$, and you can tack on $I/\sqrt{2}$ for the $\lambda$ trivial representation for a complete orthonormal basis satisfying $\text{Tr}[ {X_j^{(\lambda,\alpha)\dagger}} X_{j'}^{(\lambda',\alpha')} ] = \delta_{\lambda,\lambda'} \delta_{j,j'} \delta_{\alpha,\alpha'}$.
To see that these are tensor operators, I prefer to use the "commutator" definition of tensor operators that states that the commutator of the generators of the group $T_a$ (in $\mathcal{B(H)}$) with a tensor operator $X_j^{(\lambda,\alpha)}$ is a linear combination of tensor operators weighted by the matrix elements of the generator in representation $\lambda$.
$[T_a, X_j^{(\lambda,\alpha)}] = \sum_{j'} {T_a}^\lambda_{j'j} X_{j'}^{(\lambda,\alpha)}$
In particular, we can take linear combinations of generators in the above. I will consider the complexification of the generators $S^+ = S^x + i S^y$ from which the rest of the algebra can be defined. It is straightforward to check that one has $[S^+, X_{j=-1}^{(\lambda=adj)}] = 2 S^z = \sqrt{2} (X_{j=0}^{(\lambda=adj)})$ and $[S^+, X_{j=0}^{(\lambda=adj)}] = -\sqrt{2} S^+ = \sqrt{2} (X_{j=1}^{(\lambda=adj)})$, which coincides correctly with the adjoint representation's $(S^+)_{adj} = \begin{bmatrix}0 & \sqrt{2} & 0 \\ 0 & 0 & \sqrt{2}\\ 0 & 0 & 0 \end{bmatrix}$.
Now, in this tensor operator representation, $X_{j=-1}^{(\lambda=adj)} = S^-$ and $X_{j=1}^{(\lambda=adj)} = -S^+$ don't have the right singular values. Their singular values are not the desired $(1/\sqrt{2}, 1/\sqrt{2})$ but are instead $(1,0)$.
Uniqueness proof: One might wonder if a different choice for the operators $X_j^{(\lambda,\alpha)}$ in $\lambda = adjoint$ could work for this case of $\mathcal{H} = \mathbb{C}^2$. The answer is no, the above $\{X_j^{(\lambda,\alpha)}\}$ is the only choice for tensor operators for $\lambda = adjoint$ on this $\mathcal{H}$ (up to complex phase, which leaves singular values unchanged).
To see this, note that any choice of $\{ X_{j}^{(\lambda=adj)} \}$ has $[S^+, [S^+, X_{j=-1}^{(\lambda=adj)}]] = 2X_{j=1}^{(\lambda=adj)}$, according to the commutator definition of tensor operators above.
However, for ANY $X_{j=-1}^{(\lambda=adj)}$, $[S^+, [S^+, X_{j=-1}^{(\lambda=adj)}]] \propto = \begin{bmatrix}0 & -1\\ 0 & 0\end{bmatrix}$.
Thus, $X_{j=1}^{(\lambda=adj)} = e^{i\phi} \begin{bmatrix}0 & -1\\ 0 & 0\end{bmatrix}$. This in turn fixes $X_{j=0}^{(\lambda=adj)}$ via $X_{j=0}^{(\lambda=adj)} = \frac{1}{\sqrt{2}} [S^-, X_{j=1}^{(\lambda=adj)}]$ and fixes $X_{j=-1}^{(\lambda=adj)}$ via $X_{j=-1}^{(\lambda=adj)} = \frac{1}{2}[S^-,[S^-, X_{j=1}^{(\lambda=adj)}]]$, fixing the form of the representation up to phase $e^{i\phi}$.
Thus, all tensor operator bases of $SU(2)$ on $\mathcal{H} = \mathbb{C}^2$ have two tensor operators without the desired singular values, giving the answer of no to the question.