$\newcommand{\al}{\alpha} \newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\varepsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\EE}{\mathcal E} \newcommand{\F}{\mathcal F} \newcommand{\I}{\mathcal I} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}} \newcommand{\x}{\mathbf x} \newcommand{\size}{\text{size}} \newcommand{\pow}{\text{power}} \renewcommand{\bar}{\overline}$
It appears that in the OP's question the condition that the $X_i$ be independent copies of $X$ is missing; let us assume this condition. To make (1) meaningful and (2) nontrivial, we also need to assume that $\mu>0$. In addition, we need to define $Y_{n,i}$ on the event $\{\sum_{j=1}^n X_j\le0\}$; let then define $Y_{n,i}$ as $0$ on this event (of vanishing probability as $n\to\infty$).
Note first that (3) will fail to hold in general. E.g., suppose that $\PP(X=1)=\mu=1-\PP(X=0)$, where $\mu$ is any transcendental (non-algebraic) number in the interval $(0,1)$; for instance, one may take $\mu=e-2$. Then $\E X=\mu$ and all the nonzero values taken by $Y_{n,i}$ with nonzero probability are transcendental. Let now $f(y):=\ii{y=1}$ for all real $y$, where $\ii\cdot$ denotes the indicator. Then $\frac1n\sum_{i=1}^n f(Y_{n,i})$ is $0$ almost surely, whereas $\E f(X)=\PP(X=1)=\mu\ne0$, so that (3) fails to hold.
The positive news is that, as will be shown here, (3) will hold -- even with the almost sure (a.s.) convergence (rather than just in probability) -- if $f$ is continuous, which will be henceforth assumed. Take any positive real numbers $a$ and $\eta$. Since $f$ is continuous, it is uniformly continuous on the interval $[-2a,2a]$. So, for some $\ep\in(0,1]$ and all $x,y$ in $[-2a,2a]$ such that $|x-y|\le a\ep$, we have $|f(y)-f(x)|<\eta$. Take now any $\de>0$ such that \begin{equation*} |x-\mu|<\de\implies\big|\sqrt{\tfrac\mu x}-1\big|<\ep. \tag{4} \end{equation*}
Now, introduce events \begin{equation*} A_n:=A_{n,\de}:=\{|\bar X_n-\mu|<\de\},\quad B_i:=\{|X_i|\le a\}, \end{equation*} where $\bar X_n:=\frac1n\,\sum_{j=1}^n X_i$. By the strong law of large numbers, $\ii{A_n^c}\to0$; everywhere here ${}^c$ denotes the complement and the convergence is for $n\to\infty$; the convergence of sequences of random variables (r.v.'s) is a.s.
On the event $A_n\cap B_i$, we have $|X_i|\le a\le 2a$ and, by (4),
\begin{equation*}
|Y_{n,i}-X_i|=|X_i|\Big|\frac{\sqrt{\mu}}{\sqrt{\bar X_n}}-1\Big|\le a\ep
\end{equation*}
and hence also
$|Y_{n,i}|\le a(1+\ep)\le2a$.
So, by the mentioned uniform continuity of $f$,
\begin{equation*}
|f(Y_{n,i})-f(X_i)|<\eta\quad\text{on the event}\quad A_n\cap B_i.
\end{equation*}
Introduce now the r.v.'s \begin{equation*} Z_n:=\frac1n\sum_1^n f(Y_{n,i}),\quad T_n:=\frac1n\sum_1^n f(Y_{n,i})\ii{B_i}, \end{equation*} \begin{equation*} V_n:=\frac1n\sum_1^n f(X_i),\quad W_n:=\frac1n\sum_1^n f(X_i)\ii{B_i}. \end{equation*} Since the function $f$ was assumed to be bounded, we have $|f|\le M$ for some real $M>0$. Therefore, \begin{align*} &|Z_n-T_n|\le\frac Mn\sum_1^n \ii{B_i^c}\to M\PP(B_1^c)=M\PP(|X|>a), \\ &|T_n-T_n\ii{A_n}|=|T_n|\ii{A_n^c}\le M\ii{A_n^c}\to0, \\ &|T_n\ii{A_n}-W_n\ii{A_n}|\le \frac1n\sum_1^n |f(Y_{n,i})-f(X_i)|\ii{A_n\cap B_i}\le\eta, \\ &|W_n\ii{A_n}-W_n|=|W_n|\ii{A_n^c}\le M\ii{A_n^c}\to0, \\ &|W_n-V_n|\le\frac Mn\sum_1^n \ii{B_i^c}\to M\PP(B_1^c)=M\PP(|X|>a), \\ &V_n\to\E f(X). \end{align*} Thus, \begin{equation*} \limsup_n\Big|\frac1n\sum_1^n f(Y_{n,i})-\E f(X)\Big| =\limsup_n\Big|Z_n-\E f(X)\Big| \le2M\PP(|X|>a)+\eta. \end{equation*} Letting now $a\to\infty$ and $\eta\downarrow0$, we see that indeed a.s. \begin{equation*} \frac1n\sum_1^n f(Y_{n,i})\to\E f(X). \end{equation*}