How to prove that the Kronecker delta is the unique isotropic tensor of order 2? Is there a  way to prove that the Kronecker delta $\delta_{ij}$ is indeed the only isotropic second order tensor (i.e. invariant under rotation), i.e. so we can write $T_{ij} = \lambda \delta_{ij}$ for some constant $\lambda$?
By rotational invariance I mean:
$$
T_{ij} = T^\prime_{ij} = R_{ip} R_{jq} T_{pq}\text{,}
$$
where the matrices $R_{ij}$ are orthogonal.
It is very straightforward to show that $\delta_{ij}$ is invariant, but how can I show that it is unique?
 A: There is a simpler proof at http://www.damtp.cam.ac.uk/user/reh10/lectures/nst-mmii-chapter3.pdf, pg 61, that proceeds as follows
Starting with an isotrophic 2nd order tensor written in matrix form Tij, its coordinates are rotated 90 deg. around the 3 axis to get T'ij, then by comparing T'ij (which equals Tij as T is isotrophic) and Tij and concluding by inspection that T11 = T22, T13 = T23 = -T13 = 0, T31 = T32 = -T31 = 0, and then rotating the coordinates 90% about the 2 axis to conclude that T11 = T33 and T21 = T32 = 0 and T21 = T23 = 0.
That is, the proof is simplicity itself and is given in detail in the reference.
A: Step 1: Realize $T$ is diagonal. Let
$$R_{kl}=\begin{cases}
 -1 & \text{if }k=l=j\\
 \delta_{kl} & \text{otherwise}\\
\end{cases}$$ 
therefore $R$ is the reflection w.r.t to the hyperplane perpendicular to the j-th vector in the standard ordered basis.
$$R=R^T\land R^2=I\Rightarrow R^TR=RR^T=I$$
Therefore:
$$T_{ij}=\sum_{p,q}R_{ip}R_{jq}T_{pq}=R_{ii}R_{jj}T_{ij}\\
i\neq j\Rightarrow T_{ij}=-T_{ij}\Rightarrow T_{ij}=0$$

Step 2: Realize $T_{jj} = T_{11}$. Let
$$P_{kl}=\begin{cases}
 \delta_{jl} & \text{if } k=1\\
 \delta_{1l} & \text{if } k=j\\
 \delta_{kl} & \text{otherwise} 
\end{cases}$$
therefore $P$ is the permutation matrix that interchanges the 1st and j-th rows on left multiplication.
$$(P^TP)_{kl}=\sum_{m}P^T_{km}P_{ml}=\sum_{m}P_{mk}P_{ml}=\sum_{m\neq1,j}P_{mk}P_{ml}+\sum_{m=1,j}P_{mk}P_{ml}\\
=\sum_{m\neq1,j}\delta_{mk}\delta_{ml}+\delta_{jk}\delta_{jl}+\delta_{1k}\delta_{1l}=\sum_{m}\delta_{mk}\delta_{ml}=\delta_{kl}$$
Therefore:
$$T_{jj}=\sum_{p,q}P_{jp}P_{jq}T_{pq}=\sum_{q}P_{jq}^2T_{qq}=\sum_{q}\delta_{1q}^2T_{qq}=\sum_{q}\delta_{1q}T_{qq}=T_{11}$$
