I will show that it is not possible for $\phi=\pi/2$, so it is certainly not for general $\phi$. (actually, I don't think that it is possible for any single $\phi$ except $0$ and $\pi$, by an analogous argument).
I will multiply $A$ by $-i$ and conjugate it by
$\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & -i & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & -i \end{pmatrix}$ just to have real coefficients for simplicity, it clearly doesn't affect block-diagonalisabilty. $A$ then becomes:
$$\begin{pmatrix}
\lambda & 1 & 1 & 0 \\
1 & 0 & 0 & 0 \\
1 & 0 & -\lambda & 1 \\
0 & 0 & 1 & 0 \end{pmatrix}$$
and has characteristic polynomial $X^4-(\lambda^2+3)X^2+1$. If it was block diagonalizable, then the characteristic polynomial (seen as a polynomial with coefficients in $\mathbb{C}[\lambda]$) would factor as a product of two factors of degree $2$ in $X$:
$$X^4-(\lambda^2+3)X^2+1=(X^2+aX+b)(X^2+cX+d)$$
with $a,b,c,d\in \mathbb{C}[\lambda]$. Comparing coefficients yields:
- $bd=1$ (in particular $b,d\in \mathbb{C}$)
- $a+c=0$
- $ad+bc=a(d-b)=0$, so either $a=0$ or $b=d=\pm 1$
The case $a=0$ is impossible since then the factors would have coefficients in $\mathbb{C}$ and we couldn't get the coefficient $\lambda^2+3$. Hence the factorization must have the form
$$X^4-(\lambda^2+3)X^2+1=(X^2+aX\pm 1)(X^2-aX\pm 1)$$
Comparing coefficients of $X^2$, we have
$$-a^2\pm2=-(\lambda^2+3)$$
$$a^2\pm2=\lambda^2+3$$
$$a^2=\lambda^2+1 \ \text{or}\ a^2=\lambda^2+5$$
Since neither of $\lambda^2+1$ and $\lambda^2+5$ are squares in $\mathbb{C}[\lambda]$, we get a contradiction.
