The answer is no.

The easiest way to see this I think it to look at the concrete embedding of $SU(3)\times SU(2)\times U(1)$ in $SU(5)$. That embedding is given by

\[(A,B,\mathrm{e}^{\mathrm{i}\theta})\mapsto \begin{pmatrix}A\mathrm{e}^{\mathrm{i}\theta} & 0 \\ 0 & B\mathrm{e}^{-\mathrm{i}\theta}\end{pmatrix}\].

Given this, one can simply just compute to see that the conjugate of something of this form is not necessarily of this form. For example, the conjugate of this by

\[-\frac{1}{\sqrt{2}}\begin{pmatrix}1 & 0 & \mathrm{i} & 0 & 0 \\ \mathrm{i} & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & \mathrm{i} & 0 \\ 0 & \mathrm{i} & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & \sqrt{2}\end{pmatrix}=:\begin{pmatrix}X & Y \\ U & V\end{pmatrix}\]

is

\[\begin{pmatrix}XAX^*\mathrm{e}^{\mathrm{i}\theta}+YBY^*\mathrm{e}^{-\mathrm{i}\theta} & XAU^*\mathrm{e}^{\mathrm{i}\theta}+YBV^*\mathrm{e}^{-\mathrm{i}\theta} \\ UAX^*\mathrm{e}^{\mathrm{i}\theta}+VBY^*\mathrm{e}^{-\mathrm{i}\theta} & UAU^*\mathrm{e}^{\mathrm{i}\theta}+VBV^*\mathrm{e}^{-\mathrm{i}\theta}\end{pmatrix}\].

Taking

\[\theta :=0,\qquad A:=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & \mathrm{i} & 0 \\ \mathrm{i} & 1 & 0 \\ 0 & 0 & \sqrt{2}\end{pmatrix},\qquad B:=\frac{1}{\sqrt{2}}\begin{pmatrix}1 & \mathrm{i} \\ \mathrm{i} & 1\end{pmatrix}\],

a computation shows that this conjugate is not in the sub-group of $SU(5)$ listed above isomorphic to $SU(3)\times SU(2)\times U(1)$.