Recurrence relation complexity
Suppose that $n$ is a power of $k$, say $n = k^t$. Then
$$
T(k^t) = kT(k^{t-1}) = k^2T(k^{t-2}) = \cdots = k^tT(1) = k^t,
$$
assuming a base case of $T(1) = 1$. So for powers of $k$, we have $T(n) = n$. You can also prove that by induction: if $T(n/k) = n/k$ then $T(n) = kT(n/k) = k(n/k) = n$.