Continuity of mapping sending a function to its (brouwer) fixed point

The answer is yes. Fixed points of $f$ are zeros of the continuous $g(x)=f(x)-x$. So one has to prove the following: If $g:[0,1]^n\times[0,1]\to[0,1]^n$ is continuos, and for each $r\in[0,1]$, $g(.,r)$ has exactly one zero $x_r$ then $r\to x_r$ is continuous. Consider the set $$E=\{ (x,r):g(x,r)=0\}\subset [0,1]^{n+1}.$$ this set is closed. On the other hand, this set is the graph of the function $r\mapsto x_r$. A function on $[0,1]$ whose graph is closed is continuous. The proof is straightforward. Let $r_n\to r^*$, choose a subsequence such that $x_{r_n}\to x^*$, then $(r^*,x^*)\in E$ because $E$ is closed, and $x_{r^*}=x^*$ because there is a unique $x^*$ such that $(r^*,x^*)\in E$.

Yes, it is indeed continuous. Consider the correspondence (set-valued map) $F:[0,1]\to 2^{[0,1]^n}$ given by $$F(x)=\{y\in[0,1]^n:f(y,x)=y\}.$$ Obviously, $F$ has closed values in a compact Hausdorff space and has a closed graph. Therefore, $F$ is upper hemicontinuous. Now if $F$ is single-valued, and therefore essentally a function, being upper hemicontinuous coincides with being continuous.