You can replace $GL_1(\mathbb C)$ with its maximal compact subgroup, which is $S^1$. Since $S^1$ is an abelian compact Lie group, there is a natural $\mathbb Z/n$-equivariant equivalence $$B_{\mathbb Z/n}S^1\xrightarrow{\simeq} \mbox{Map}(E\mathbb Z/n, BS^1).$$

See, for example, this MO question for a discussion of this equivalence, with references and some generalizations.

It follows that for a compact $\mathbb Z/n$-complex $X$, $\mbox{Vect}^1_{\mathbb Z/n}(X)$ is $\pi_0$ of the space $$\mbox{Map}(X\times E\mathbb Z/n, K(\mathbb Z, 2))^{\mathbb Z/n}\simeq \mbox{Map}(X\times_{\mathbb Z/n} E\mathbb Z/n, K(\mathbb Z, 2))$$

So the answer is the second cohomology group of the homotopy orbit space $X\times_{\mathbb Z/n} E\mathbb Z/n$, rather than a Bredon cohomology group of $X$ with coefficients in a Mackey functor.

You can replace $GL_1(\mathbb C)$ with its maximal compact subgroup, which is $S^1$. Since $S^1$ is an abelian compact Lie group, there is a natural $\mathbb Z/n$-equivariant equivalence $$B_{\mathbb Z/n}S^1\xrightarrow{\simeq} \mbox{Map}(E\mathbb Z/n, BS^1).$$

See, for example, this MO question for a discussion of this equivalence, with references and some generalizations.

It follows that for a compact $\mathbb Z/n$-complex $X$, $\mbox{Vect}^1_{\mathbb Z/n}(X)$ is $\pi_0$ of the space $$\mbox{Map}(X\times E\mathbb Z/n, K(\mathbb Z, 2))^{\mathbb Z/n}\simeq \mbox{Map}(X\times_{\mathbb Z/n} E\mathbb Z/n, K(\mathbb Z, 2))$$

So the answer is the second cohomology group of the homotopy orbit space $X\times_{\mathbb Z/n} E\mathbb Z/n$, rather than a Bredon cohomology group of $X$ with coefficients in a Mackey functor.

The Leray-Serre spectral sequence for computing the (co)homology of the homotopy orbit space is often tractable.