Let \(\mathbf x=\{x_i~:~1\le i\le n\}\) be the sample, with \(n\) the size of the sample.
The mean of the sample is
\(\displaystyle\bar x=\frac1n\sum_{i=1}^nx_i\)
Let \(\mathbf y\) be the same data set but with each value scaled up by 5. Then the mean here is
\(\bar y=\displaystyle\frac1n\sum_{i=1}^ny_i=\frac5n\sum_{i=1}^nx_i=5\bar x\)
The standard deviation for \(\mathbf x\) is given by
\(s_{\mathbf x}=\sqrt{\displaystyle\frac1{n-1}\sum_{i=1}^n(\bar x-x_i)^2}\)
For \(\mathbf y\), you would have
\(s_{\mathbf y}=\sqrt{\displaystyle\frac1{n-1}\sum_{i=1}^n(\bar y-y_i)^2}\)
\(s_{\mathbf y}=\sqrt{\displaystyle\frac1{n-1}\sum_{i=1}^n(5\bar x-5x_i)^2}\)
\(s_{\mathbf y}=\sqrt{\displaystyle\frac{25}{n-1}\sum_{i=1}^n(\bar x-x_i)^2}\)
\(s_{\mathbf y}=5\sqrt{\displaystyle\frac1{n-1}\sum_{i=1}^n(\bar x-x_i)^2}\)
\(s_{\mathbf y}=5s_{\mathbf x}\)
