For all $x\in SU(n)$ there is a $y\in SU(n)$ such that $b=yxy^{-1}$ is a matrix in block form, where each block is an element of $SU(1)$ or $SU(2)$
I know that every unitary matrix is normal and therefore diagonalizable, therefore, given $A\in SU(n)$ there is $S\in U(n)$ and diagonal matrix $D$ such that $A=SDS^*$. But I don't know how to use this to prove what I want. I'm trying to reason by induction, what can I do? Thank you.
It is true that every element of $SO(n)$ is conjugate to a block-diagonal matrices with blocks from $SO(2)$, and possibly one $1\times 1$ block $[1]$ at the end (so, in $SO(1)$). This is useful for showing $\exp:\mathfrak{so}(n)\to SO(n)$ is an onto function. However, it is not true that every element of $SU(n)$ is conjugate to a block-diagonal matrix with blocks from $SU(2)$ and $SU(1)$. This is because $SU(1)$ is trivial, and any diagonal entry $1$ would entail fixed points, so according to this statement if an element of $SU(n)$ had no fixed points the dimension $n$ would have to be even (since it is comprised of $2\times 2$ blocks). A quick counterexample is $\omega I_3\in SU(3)$, where we have $I_3$ the $3\times 3$ identity matrix and $\omega$ either nontrivial cube root of unity.
What is true instead is that every element of $SU(n)$ is conjugate to a diagonal matrix. Since the matrix is still unitary, its entries must be unit complex numbers ("phasors"), which are elements of $U(1)$. Moreover the determinant is $1$, so if the diagonal matrix is $\mathrm{diag}(e^{i\theta_1},\cdots,e^{i\theta_n})$ the determinant is $e^{i(\theta_1+\cdots+\theta_n)}$, which implies $\theta_1+\cdots+\theta_n$ is an integer multiple of $2\pi$ (subtracting that integer multiple from one of $\theta_i$ does not change the diagonal matrix, so without loss of generality $\theta_1+\cdots+\theta_n=0$).
The map $\exp:\mathfrak{su}(n)\to SU(n)$ intertwines with the action of $SU(n)$ (by conjugation, which is called the adjoint representation of $SU(n)$ on $\mathfrak{su}(n)$). That is, this holds for all $g\in SU(n),X\in\mathfrak{su}(n)$:
$$ \exp(gXg^{-1})=g\exp(X)g^{-1}. $$
(Can you verify this using the power series formula?) This actually holds more generally for Lie groups $G$, although the equation above only makes sense for matrix Lie groups as written.
It suffices to show the aforementioned diagonal matrices are in the range of $\exp$. But
$$ \mathrm{diag}(e^{i\theta_1},\cdots,e^{i\theta_n})=\exp \mathrm{diag}(i\theta_1,\cdots,i\theta_n) $$
and $\mathrm{diag}(i\theta_1,\cdots,i\theta_n)$ is in $\mathfrak{su}(n)$ (verify this).
The fact that elements of $SU(n)$ or $\mathfrak{su}(n)$ are conjugate to diagonal matrices in $U(1)$ or $\mathfrak{u}(1)$ respectively follows from the spectral theorem for normal operators. Note that if two elements of $U(n)$ are conjugate, then they are also conjugate by an element of $SU(n)$. If we write $A=SDS^{\ast}$ with $A,S,D\in U(n)$ (notice then that $S^{\ast}=S^{-1}$ so this is conjugacy in the group-theoretic sense) then we can scale $S$ to $R=(\det S)^{-1/n}S$ (where we pick any $n$th root) so that $R\in SU(n)$ but also $A=RDR^{-1}$ too.