Let $x^{+} : \{0, \dotsc, dN-1 \} \to \mathbb{C}$ be a concatenation of $d$ copies of a signal $x: \{0, \dotsc, N-1\} \to \mathbb{C}$. Then $x^{+}$ is periodic with period $N$ ($\bmod$ $dN$). Denote $e^{2\pi i/n}$ by $\zeta_n$ for any $n$. Direct computation of the (unnormalized) Fourier transform gives
$ \widehat{x^+}(m) = \sum_{k=0}^{dN-1}x_k^+ \zeta_{dN}^{-mk} = \sum_{k = 0}^{N-1}\left( x_k \sum_{j = 0}^{d-1} \zeta_{dN}^{-m(k + jN)}\right) = \sum_{k = 0}^{N-1}\left( x_k \zeta_{dN}^{-mk}\sum_{j = 0}^{d-1} \zeta_{d}^{-mj}\right). $
The inner sum simplifies to
$ \sum_{j = 0}^{d-1}\zeta_d^{-mj} = \begin{cases} d & \textrm{if } d \textrm{ divides } m\\[1em] 0 & \textrm{otherwise} \end{cases}. $
Plug this in to get
$ \widehat{x^+}(m) = \begin{cases} d \cdot \widehat{x}(\tfrac{m}{d}) & \textrm{if } d \textrm{ divides } m\\[1em] 0 & \textrm{otherwise} \end{cases}. $
This shows that $\widehat{x^+}$ is a multiple of a spread out version of $\widehat{x}$ with a bunch of zeroes added in between.