Show that similar matrices have same trace

Hint: By very definition, two matrices $A,B$ are similar iff there exists an invertible matrix $S$ such that $A=SBS^{-1}$. Now apply the trace on both sides, and conclude using associativity of the matrix product.

You can use the characteristic polynomials. If $A$ is similar to $B$, then their characteristic polynomials $f_A(x)$ and $f_B(x)$ are identical. Since $$ f_A(x)=x^n-tr(A)x^{n-1}+\ldots\pm\det(A)\\ $$ and $$ f_B(x)=x^n-tr(B)x^{n-1}+\ldots\pm\det(B) $$ it follows in particular that $tr(A)=tr(B)$.

Suppose that $A$ is similar to $B$. Then there exist $C$ such that: $B=CAC^{-1}$. So: $tr(B)=tr(CAC^{-1})$, now let me call $X=CA$, and $Y=C^{-1}$, then $tr(B)=tr(XY)=tr(YX)=tr(C^{-1}CA)=tr(A)$.