Extending from If $f$ is strictly convex in a convex set, show it has no more than 1 minimum and Examples of $f$ strictly convex, either with one minimizer or with no minimizer.
Can we have the same minimisers $(x_1^*,x_2^*,\cdots,x_n^*)$ for $$f\left( \frac{1}{n} \sum_{j=1}^{n} x_j\right)$$ and $$ \frac{1}{n} \sum_{j=1}^{n} f\left( x_j\right)?$$ where $f:\mathbb{R}^+ \to \mathbb{R}^+$ is strictly convex and $n$ is finite natural number.
If you're asking whether the two quantities $$f\left( \frac{1}{n} \sum_{j=1}^{n} x_j\right)\tag1$$ and $$\frac{1}{n} \sum_{j=1}^{n} f\left( x_j\right)?\tag2$$ can have the same minimizers, the answer is yes: Let $a$ be the minimizer of $f$, and set $x_1=x_2=\cdots=x_n=a$.
If you're asking whether the two quantities (1) and (2) must have the same minimizers, the answer is no. The quantity (1) is minimized by any $x_1,\ldots,x_n$ for which $\sum x_i= na$, whereas $x_1=x_2=\cdots=x_n=a$ is the only minimizer for (2). If $a>0$, then the second solution set is a proper subset of the first.