Answer
False.
Work Step by Step
If $c$ is the period of $f(t)$, then the period of $f(\frac{1}{2}t)$ is $2c$
$f(\frac{1}{2}[t+c])=f(\frac{1}{2}t+\frac{1}{2}c)\ne f(\frac{1}{2}t)$, because $\frac{1}{2}c$ is not the period.
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