Suppose $X$ and $Y$ are two random variables that have the same distribution. Does
$$P[X \leq t \mid Y=a]$$
be necessarily equal to $$\;\ P[Y \leq t \mid X=a]?$$ Note that if $X$ and $Y$ are bivariate normal with correlation $\rho$ and each is marginally $N(\mu,\sigma^2)$, then it is necessarily true because both conditional distributions would be
$$N(\mu+\rho(a-\mu),(1-\rho^2)\sigma^2).$$
A simple counterexample:
$$ P(X = 1, Y = 2) = \frac{1}{3}\\ P(X = 2, Y = 3) = \frac{1}{3}\\ P(X = 3, Y = 1) = \frac{1}{3} $$
Then $P(X \le 1 | Y = 2) = 1$, but $P(Y \le 1 | X = 2) = 0$.
There are already answers with simple examples, so why one more? Because it is interesting to look at the general pattern. The wanted property is a kind of symmetry, so we should look for some asymmetrical joint distribution for $(X, Y)$. Si if $(X, Y)$ has a permutable distribution in the sense that $(X,Y)$ and $(Y,X)$ have the same joint distribution, so that for the joint cumulative we have $F(x,y)=F(y,x)$ for all $(x,y)$, then the sought-after property will hold.
Let us use copulas. Let $F$ be the joint cdf (cumulative distribution function) and $$ \DeclareMathOperator{\P}{\mathbb{P}} C(u,v)= \P(F(x) \le u, F(Y)\le v) $$ By the Fréchet–Hoeffding copula bounds (see linked wiki article above) we have $$ W(u,v) \le C(u,v) \le M(u,v) $$ where $W(u,v)= \max(u+v-1,0)$ and $M(u,v)=\min(u,v)$. Both $W,M$ are copulas. $W$ describes the anti-monotonic case $X=U, Y=1-U$ for $U$ some uniform random variable. Now you can check that $W$ gives a counterexample. $M$ corresponds to $X=U, Y=U$ which is not a counterexample.
How about $X=-Y = \begin{cases} 0 \\ 1 \end{cases}\quad$ each with probability $1/2.$
If you have a bivariate normal distribution $X,Y$ and change the sign of each $Y$ according to the sign of $(X - \lfloor X \rfloor - 0.5)$, then you don't change the marginal distribution of $Y$. This is because $(X - \lfloor X \rfloor - 0.5)$ is with equal probability negative and positive. However, the joint and conditional distribution is changed a lot in an asymmetric way.
n = 1000
x = rnorm(n)
y = rnorm(n)
y = sign(x-floor(x)-0.5)*abs(y)
plot(x,y,pch=20, cex = 0.6, xlim = c(-4,4), ylim = c(-4,4))
Not necessarily true. Let $X$ and $Y$ be discrete random variables that take the values in {1, 2, 3} each with probability $\frac{1}3$; i.e. they have a discrete uniform distribution. Consider the joint probability mass function represented by the matrix below where the element in row $i$ and column $j$ is $P[X=i, Y=j]$:
$$\frac{1}{60}\begin{bmatrix} 3 & 6 & 11 \\ 3 & 12 & 5 \\ 14 & 2 & 4 \end{bmatrix}$$
Note that all rows and columns add up to $\frac{1}3$ and therefore the marginal distributions are the discrete uniform as stated. Now calculate the conditional probability $$P[X\leq2\mid Y=1]=\frac{\frac{3}{60}+\frac{3}{60}}{\frac{3}{60}+\frac{3}{60}+\frac{14}{60}}=\frac{3}{10}.$$ On the other hand, $$P[Y\leq2\mid X=1]=\frac{\frac{3}{60}+\frac{6}{60}}{\frac{3}{60}+\frac{6}{60}+\frac{11}{60}}=\frac{9}{20}.$$