Let $ R $ and $ S $ be two commutative rings and let $ f\colon R\to S $ be a homomorphism of rings. For any $ S $-module $ N $ denote with $ f^*N $ the module obtained from $ N $ by restriction of scalars ($ r\cdot_R n = f(r)n $).
Now, given $ N $ one can obtain an $ R $-module in the two following ways: first take the exterior power of $ N $ and then immediately restrict the scalars, $ f^*\bigwedge_S^k N $; or first restrict the scalars and then take the exterior power, $ \bigwedge_R^k f^*N $.
Is there any relationship between $ f^*\bigwedge_S^k N $ and $ \bigwedge_R^k f^*N $? Are they isomorphic?
Ok as suggested by @coiso and @Martin Brandenburg I'll look at a concrete example: $ \iota\colon \mathbb R\hookrightarrow \mathbb C $. Then $$ \dim_{\mathbb R}\iota^*\mathbb C^n = 2n\quad \leadsto\quad \dim_{\mathbb R}\bigwedge\nolimits_{\mathbb R}^k \iota^*\mathbb C^n = \binom{2n}{k} $$ but $$ \dim_{\mathbb C}C^n = n\text{,}\, \dim_{\mathbb C}\bigwedge\nolimits_{\mathbb C}^k\mathbb C^n = \binom{n}{k}\quad\leadsto \quad \dim_{\mathbb R}\iota^*\bigwedge\nolimits_{\mathbb C}^k\mathbb C^n = 2\binom{n}{k} $$ and in general they are different.
I still have to work out the natural transformation $$ \bigwedge\nolimits_R^kf^\ast\to f^\ast\bigwedge\nolimits^k_S $$ but I hope I could do it. Sorry for the rush!