Let $M$ be positive operator on a complex Hilbert space $E$.
Since the square root of a positive operator is unique, then
$$\begin{pmatrix} M^{1/2} & \\ & \ddots & \\ & & M^{1/2} \end{pmatrix} =\begin{pmatrix} M & \\ & \ddots & \\ & & M \end{pmatrix}^{1/2} .$$
Now if $t\in (0,\infty)$ be real number. Is $$\begin{pmatrix} M^{t} & \\ & \ddots & \\ & & M^{t} \end{pmatrix} =\begin{pmatrix}M & \\ & \ddots & \\ & & M \end{pmatrix}^{t} ?$$
If $A$, $B$ are unital $C^\ast$-algebras and $\pi\colon A\to B$ is a unital $\ast$-homomorphism, then $\pi(f(x))=f(\pi(x))$ for every normal $x\in A$ and continuous function $f$ on the spectrum of $x$. If $f$ is a polynomial in $z$ and $\bar z$, this follows easily from the properties of a unital $\ast$-homomorphism, and for general continuous $f$ one can approximate by polynomials in $z$ and $\bar z$.
Now you can apply this result to the case when $B=M_n(A)$, $\pi(x)=\mathrm{diag}(x,\dots,x)$ and $f(\lambda)=\lambda^t$.